A TASTE OF MATHEMATICS
´ AIME–T–ON LES MATHEMATIQUES
Volume / Tome VII THE MURRA MURRAY KLAMKIN KLAMKIN PROBLEM PROBLEMS S
CANADIAN COLLECTION — Part 1.
Edited by
Andy Liu University of Alberta and
Bruce Shawyer Memorial University of Newfoundland
Published by the Canadian Mathematical Society, Ottawa, Ontario and produced by the CMS ATOM Office, St. John’s, NL Publi´ Pub li´e par la Soci´ Soc i´et´ et´e math´ematiqu ema tiquee du Canada, Cana da, Ottawa (Ontario (Onta rio)) et produit par le Bureau ATOM de la SMC, St. John’s, NL
Printed in Canada by / imprim´e au Canada par The University of Toronto Press Incorporated ISBN 0-919558-16-X
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c 2005
Canadi Can adian an Mathe M athemati matical cal Society So ciety / Soci´ So ci´et´ et´e math´ ma th´ematique emat ique du Canada Cana da
A TASTE OF MATHEMATICS
´ AIME–T–ON LES MATHEMATIQUES
Volume / Tome VII THE MURRA MURRAY KLAMKIN KLAMKIN PROBLEM PROBLEMS S
CANADIAN COLLECTION — Part 1.
Edited by
Andy Liu University of Alberta and
Bruce Shawyer Memorial University of Newfoundland
The ATOM series The booklets in the series, A Taste of Mathematics, are published by the Canadian Mathematical Society (CMS). They are designed as enrichment materials for high school students with an interest in and aptitude for mathematics. Some booklets in the series will also cover the materials useful for mathematical competitions at national and international levels.
La collection ATOM Publi´es par la Soci´et´e math´ematique du Canada (SMC), les livrets de la collection Aime-t-on les math´ ematiques (ATOM) sont destin´es au perfectionnement des ´etudiants du cycle secondaire qui manifestent un int´ erˆet et des aptitudes pour les math´ematiques. Certains livrets de la collection ATOM servent ´egalement de mat´eriel de pr´eparation aux concours de math´ematiques sur l’´echiquier national et international.
Editorial Board / Conseil de r´ edaction Editor-in-Chief / R´edacteur-en-chef Bruce Shawyer Memorial University of Newfoundland / Universit´e Memorial de Terre-Neuve
Associate Editors / R´edacteurs associ´es Edward J. Barbeau University of Toronto / Universit´e de Toronto
Malgorzata Dubiel Simon Fraser University / Universit´e Simon Fraser
Joseph Khoury University of Ottawa / Universit´e d’Ottawa
Antony Thompson Dalhousie University / Universit´e Dalhousie
Managing Editor / R´edacteur-g´erant Graham P. Wright University of Ottawa / Universit´e d’Ottawa
iv
Table of Contents Preface
1
1 Quickies
2
2 Combinatorics and Number Theory
27
3 Functions and Polynomials
39
4 Expressions and Identities
51
v
QUICKIES Murray Klamkin was famous for his Quickies, problems that had quick and neat solutions. We present all the Quickies published in CRUX MATHEMATICORUM, with some slight editing by Bruce Shawyer.
PROBLEMS AND SOLUTIONS The problems have been selected by Andy Liu and arranged into sets according to topic. The solutions are as published in CRUX MATHEMATICORUM, with some slight editing by Bruce Shawyer. Solutions from pre-LATEX editions were coded by students at Memorial University of Newfoundland, funded by the Canadian Mathematical Society. Special thanks are due to Karelyn Davis, Alyson Ford, Don Hender, Shawna Gammon, Paul Marshall, Shannon Sullivan and Rebecca White. These problems exhibit the special talents of Murray Klamkin. They cover a very wide range of topics, and show a great deal of insight into what is possible in there areas. They demonstrate that Murray Klamin was a problem setter par excellence . We are very greatful to have known him and to have been inspired by him. Problem numbers and references are to [year : page number] are as in CRUX MATHEMATICORUM. When a problem number is followed by a star , this means that the problem was proposed without a solution.
Andy Liu Department of Mathematics University of Alberta Edmonton Alberta
Bruce Shawyer Department of Mathematics Memorial University of Newfoundland St. John’s Newfoundland and Labrador
1
Problems posed by Murray Klamkin. Unless otherwise stated, these problems were proposed by Murray Klamkin alone. Quickies Part 1 1—54 Combinatorics and Number Theory Part 1 429, 1456, 2054, 1863, 1027, 969, 1561, 2034, 1752, 1434. Vectors and Matrices Part 4 1200, 1721, 1482, 1693, 2005, 398, 1314, 3024, 1207, 1242. Functions and Polynomials Part 1 299, 254, 1423, 2014, 1110, 1283. Expressions and Identities. Part 1 1304, 287, 1594, 1522, 830, 1996, 1362. Numerical Approximations Part 4 1003, 1213, 1371. Algebraic Inequalities. Part 2 347, 1642, 2615, 1703, 1734, 1445, 2064, 2095, 2044, 1652, 1742, 1674, 1774, 323, 805, 1394, 2734, 2839, 1662. Trigonometric Inequalities. Part 3 1414, 908, 1352, 1712, 1542, 1613, 1503, 2084, 1332, 1801, 1271, 1060, 958, 1962. Geometric Inequalities. Part 4 1165, 1473, 1574, 1764, 1296, 506, 1131, 1985, 1945. The Triangle. Part 3 1872, 1605, 1385, 2848, 210, 1076, 1532, 2618. Cevian Lines. Part 4 2614, 548, 485, 2613, 1621, 1631. Central Symmetry. Part 3 1062, 1348, 1513. Conic Sections. Part 2 2616, 1975, 1935, 1405, 520. Solid Geometry. Part 2 375, 1784, 1553, 1581, 330, 478, 2617, 1261. Higher Dimensions. Part 2 2651, 224, 1086, 1465, 2733, 2024, 1793. Calculus. Part 4 1178, 1494, 1322, 1147, 273. Problems dedicated to Murray Klamkin Part 4 1241, 2619, 2620, 2621 Klamkin Problems of September 2005 Part 4 K–01 through K–15
2
1
Quickies
1. Determine the extreme values of r 1 /h1 + r2 /h2 + r3 /h3 + r4 /h4 where h 1 , h 2, h3 , h4 are the four altitudes of a given tetrahedron T and r1 , r2 , r3 , r4 are the corresponding signed perpendicular distances from any point in the space of T to the faces.
2. Determine the minimum value of the product P = (1 + x1 + y1 )(1 + x2 + y2 ) . . . (1 + xn + yn )
≥ 0, and x 1 x2 . . . xn = y1y2 . . . yn = an.
where x i , yi
3. Prove that if F (x,y,z) is a concave function of x, y, z, then {F (x,y,z)}−2 is a convex function of x, y , z.
4.
If a, b, c are sides of a given triangle of perimeter p, determine the maximum values of
− b)2 + (b − c)2 + (c − a)2, (ii) |a − b| + |b − c| + |c − a|, (iii) |a − b||b − c| + |b − c||c − a| + |c − a||a − b|. (i) (a
5. If A, B, C are three dihedral angles of a trihedral angle, show that sin A, sin B, sin C satisfy the triangle inequality.
6. Are there any integral solutions (x,y,z) of the Diophantine equation (x
− y − z)3 = 27xyz
other than ( a,a,a) or such that xy z = 0?
−
7. Does the Diophantine equation (x
− y − z)(x − y + z)(x + y − z) = 8xyz
have an infinite number of relatively prime solutions?
8.
It is an easy result using calculus that if a polynomial P (x) is divisible by its derivative P (x), then P (x) must be of the form a(x r)n . Starting from the known result that P (x) 1 = P (x) x ri
−
−
where the sum is over all the zeros ri of P (x) counting multiplicities, give a non-calculus proof of the above result.
9. Solve the simultaneous equations x2 (y + z) = 1,
y 2 (z + x) = 8,
z 2 (x + y) = 13.
3
10. Determine the area of a triangle of sides a, b, c and semiperimeter s if (s
− b)(s − c) = ha ,
(s
− c)(s − a) = kb ,
(s
− a)(s − b) = cl ,
where h, k, l are consistent given constants.
11. Prove that 3(x2 y + y 2 z + z 2 x)(xy 2 + yz 2 + zx2) where x, y, z
12.
≥ xyz(x + y + z)2
≥ 0.
Determine all integral solutions of the Diophantine equation (x8 + y 8 + z 8 ) = 2(x16 + y 16 + z 16 ).
13. Determine all the roots of the quintic equation 31x5 + 165x4 + 310x3 + 330x2 + 155x + 33 = 0.
14.
If F (x) and G(x) are polynomials with integer coefficients such that F (k)/G(k) is an integer for k = 1, 2, 3, . . ., prove that G(x) divides F (x).
15. Given that ABCDEF is a skew hexagon such that each pair of opposite sides are equal and parallel. Prove that the midpoints of the six sides are coplanar.
16.
If a, b, c, d are the lengths of sides of a quadrilateral, show that
√ a √ , (4 + a)
√ b √ , (4 + b)
√ d √ , (4 + d)
√ c √ , (4 + c)
are possible lengths of sides of another quadrilateral.
17.
Determine the maximum value of the sum of the cosines of the six dihedral angles of a tetrahedron.
18. Which is larger
√ 3
( 2
− 1)1/3
or
19. Prove that 3
× min
− 3
a b c b c a + + , + + b c a a b c
1/9
≥
3
2/9 +
(a + b + c)
3
4/9?
1 1 1 + + a b c
where a, b, c are sides of a triangle.
20. Let ω = e iπ/13 . Express 1−1 ω as a polynomial in ω with integral coefficients. 21. Determine all integral solutions of the simultaneous Diophantine equations x2 + y 2 + z 2 = 2w2 and x 4 + y 4 + z 4 = 2w4 .
4
22.
Prove that if the line joining the incentre to the centroid of a triangle is parallel to one of the sides of the triangle, then the sides are in arithmetic progression and, conversely, if the sides of a triangle are in arithmetic progression then the line joining the incentre to the centroid is parallel to one of the sides of the triangle.
23. Determine integral solutions of the Diophantine equation x y y z z w w x + + + =0 x + y y + z z + w w + x
−
−
−
−
(joint problem with Emeric Deutsch, Polytechnic University of Brooklyn).
24. For x, y , z > 0, prove that 1 (i) 1 + (x + 1)
1 1+ x(x + 2)
≥
x
,
(ii) [(x + y)(x + z)]x [(y + z)(y + x)]y [(z + x)(z + y)]z
25.
≥ [4xy]x[4yz]y [4zx]z .
If ABCD is a quadrilateral inscribed in a circle, prove that the four lines joining each vertex to the nine point centre of the triangle formed by the other three vertices are concurrent.
26. How many six digit perfect squares are there each having the property that if each digit is increased by one, the resulting number is also a perfect square?
27. Let V i W i , i = 1, 2, 3, 4, denote four cevians of a tetrahedron V 1 V 2 V 3 V 4 which are concurrent at an interior point P of the tetrahedron. Prove that
≤ max V iW i ≤ longest edge.
P W 1 + P W 2 + P W 3 + P W 4
28.
Determine the radius r of a circle inscribed in a given quadrilateral if the lengths of successive tangents from the vertices of the quadrilateral to the circle are a, a, b, b, c, c, d, d, respectively.
29. Determine the four roots of the equation x 4 + 16x − 12 = 0. 30. Prove that the smallest regular n–gon which can be inscribed in a given regular n–gon is one whose vertices are the mid-points of the sides of the given regular n–gon.
31. If 311995 divides a 2 + b2, prove that 311996 divides ab. 32. Determine the minimum value of S =
( + 1) + 2( − 2) + ( + 3) + ( + 1) + 2( − 2) + ( + 3) ) + ( + 1) + 2( − 2) + ( + 3) + ( + 1) + 2( − 2) + ( + 3) a
c
2
2
b
d
2
2
c
a
2
b
2
d
2
2
c
2
a
2
d
2
b
2
where a, b, c, d are any real numbers.
33. A set of 500 real numbers is such that any number in the set is greater than one-fifth the sum of all the other numbers in the set. Determine the least number of negative numbers in the set.
5
34. Prove that a+b+c
≥
b2 + c2
− a2 +
c2 + a2
− b2 +
where a, b, c are sides of a non-obtuse triangle.
a2 + b2
− c2 ,
35. Determine the extreme values of the area of a triangle ABC given the lengths of the two altitudes h a , h b and the side BC = a.
36.
Determine the maximum area of a triangle AB C given the perimeter p and the angle A.
37. Determine the minimum value of
a2 + a3 + a4 + a5 a1
1/2
where the sum is cyclic over the positive numbers a 1 , a2 , a3 , a 4 , a 5 .
38. ABCD and AB C D are any two parallelograms in a plane with A opposite to C and C . Prove that B B , C C and DD are possible sides of a triangle.
39. Determine the maximum value of S = 4(a4 + b4 + c4 + d4 ) where 1
40.
≥ a, b, c, d ≥ 0.
If a, b, c, d are > 0, prove or disprove the two inequalities:
ab bc cd da c + d + a + b (ii) a 2 b + b2 c + c2 d +
(i)
− (a2 bc + b2cd + c2da + d2ab) − (a2b + b2c + c2d + d2 a) ,
≥
d2 a
a + b + c + d,
≥
abc + bcd + cda + dab.
41. Determine all the points P (x,y,z), if any, such that all the points of tangency of the enveloping (tangent) cone from P to the ellipsoid b > c), are coplanar.
x2 a2
+
y2 b2
+
z2 c2
= 1 (a >
42. Determine whether or not there exists a set of 777 distinct positive integers such that for every seven of them, their product is divisible by their sum.
43.
If R is any non-negative rational approximation to better rational approximation.
√ 5, determine an always
44.
A sphere of radius R is tangent to each of three concurrent mutually orthogonal lines. Determine the distance D between the point of concurrence and the centre of the sphere.
45.
If P (x , y , z , t) is a polynomial in x, y, z, t such that P (x,y,z,t) = 0 for all real x, y , z, t satisfying x 2 + y 2 + z 2 t2 = 0, prove that P (x , y , z , t) is divisible by
−
x2 + y 2 + z 2
− t2 .
6
46. From a variable point P on a diameter AB of a given circle of radius r, two segments P Q and P R are drawn terminating on the circle such that the angles QP A and RP B are equal to a given angle θ. Determine the maximum length of the chord QR. (rs)! is an integer, where r, s are positive integers, prove that 47. Using that s!(r!) s (rst)! is an integer for positive integers r, s, t. t!(s!)t (r!)ts y) 48. Determine the range of tan(x + tan x sin y =
given that
√
2 sin(2x + y) .
49.
A, B , C are acute angles such that sin 2 A + sin2 B + sin2 C = 2. Prove that A + B + C < 180 ◦ .
50. Determine the maximum and minimum values of
a2 cos2 θ + b2 sin2 θ +
where a and b are given constants.
a2 sin2 θ + b2 cos2 θ ,
x for x in [0, 2π]. 51. Determine the range of values of tan3 tan x 52. How large can the sum of the angles of a spherical right triangle be? 53. Let ABC be a spherical triangle whose mid-points of the sides are A , B,
and C . If B C is a quadrant
π
2
, find the maximum value of A B + A C .
54. Let ABC be a triangle with centroid G. Determine the point P in the plane of ABC such that AP AG + BP BG + CP CG is a minimum and express this minimum value in terms of the side lengths of ABC .
·
·
·
7
KLAMKIN QUICKIES SOLUTIONS
1. Determine the extreme values of r 1 /h1 + r2 /h2 + r3 /h3 + r4 /h4 where h 1 , h 2, h3 , h4 are the four altitudes of a given tetrahedron T and r1 , r2 , r3 , r4 are the corresponding signed perpendicular distances from any point in the space of T to the faces. Solution . If the face areas and volume of the tetrahedron are F 1 , F 2 , F 3 , F 4 , and V respectively, then r1 F 1 + r2 F 2 + r3 F 3 + r4 F 4 = 3V, and h 1 F 1 = h2 F 2 = h 3 F 3 = h 4 F 4 = 3V . Now eliminating the F i ’s, we get r1 /h1 + r2 /h2 + r3 /h3 + r4 /h4 = 1 (a constant).
2. Determine the minimum value of the product P = (1 + x1 + y1 )(1 + x2 + y2 ) . . . (1 + xn + yn )
≥ 0, and x 1 x2 . . . xn = y1y2 . . . yn = an.
where x i , yi
Solution . More generally, consider P = (1 + x1 + y1 +
· · · + w1)(1 + x2 + y2 + · ·· + w2) . . . (1 + xn + yn + ·· · + wn)
where x1 x2 . . . xn = ξ n , y1 y2 . . . yn = ηn , . . . , w1 w2 . . . wn xi , yi , . . . , wi 0. Then by H¨older’s inequality,
=
ω n , and
≥
P 1/n
≥
or
1+
/n
x1i
+
/n
yi1
+...+
/n
wi1
≥ (1 + ξ + η + . . . + ω)n.
P In this case ξ = η = a, so
≥ (1 + 2a)n.
P
3. Prove that if F (x,y,z) is a concave function of x, y, z, then {F (x,y,z)}−2 is a convex function of x, y , z .
Solution . More generally G(F ) is a convex function where G is a convex decreasing function. By convexity of G, λG F (x1 , y1 , z1) +(1 λ)G F (x2 , y2 , z2 )
{
} −
{
} ≥ G{λF (x1, y1, z1)+(1−λ)F (x2 , y2, z2)}.
By concavity of F , λF (x1 , y1 , z1 )+(1 λ)F (x2 , y2 , z2 )
−
≤ F ([λx1 +(1−λ)x2], [λy1+(1−λ)y2], [λz1+(1−λ)z2]).
Finally, since G is decreasing, λG F (x1 , y1 , z1) + (1
{
}
− λ)G{F (x2 , y2, z2)} ≥ G{F ([λx1 + (1 − λ)x2 ], [λy1 + (1 − λ)y2 ], [λz1 + (1 − λ)z2 ])}.
8
More generally and more precisely, we have the following known result: if F (X ) is a concave function of X = (x1 , x2 , . . . , xn ) and G(y) is a convex decreasing function of y where y is a real variable and the domain of G contains the range of F , then G F (X ) is a convex function of X .
{
4.
}
If a, b, c are sides of a given triangle of perimeter p, determine the maximum values of
− b)2 + (b − c)2 + (c − a)2, (ii) |a − b| + |b − c| + |c − a|, (iii) |a − b||b − c| + |b − c||c − a| + |c − a||a − b|. Solution . (i) (a − b)2 + (b − c)2 + (c − a)2 = 2( a2 − bc) ≤ kp2 . Let c = 0, so that k ≥ 1/2. We now show that k = 1/2 suffices. Here, (i) (a
2( reduces to
− a2
bc)
2bc + 2ca + 2ab
≤ 21 (a + b + c)2
− a2 − b2 − c2 ≥ 0.
√ a, √ b, √ c or √ √ √ √ √ √ √ √ √ √ √ √ ( a + b + c)( a + b − c)( a − b + c)(− a + b + c).
The LHS is 16 times the square of the area of a triangle of sides
There is equality iff the triangle is degenerate with one side 0. (ii) a
| − b| + |b − c| + |c − a| ≤ kp. Letting c = 0, k ≥ 1. To show that k = 1 suffices, assume that a ≥ b ≥ c, so that |a − b| + |b − c| + |c − a| = 2a − 2c ≤ a + b + c and there is equality iff c = 0.
| − b| |b − c| + |b − c||c − a| + |c − a||a − b| ≤ kp 2. Letting c = 0, k ≥ 1/4. To show that k = 1/4 suffices, let a = y + z, b = z + x, c = x + y where z ≥ y ≥ x ≥ 0. Our inequality then becomes |x − y| |z − y| + |y − z| |z − x| + |z − x||x − y| ≤ (x + y + z)2 (iii) a
or x2
− y2 + z2 + yz − 3zx + xy ≤ x2 + y2 + z2 + 2yz + 2zx + 2xy
or 2y 2 + 5zx + 1xy + 1yz
≥ 0.
There is equality iff x = y = 0 or equivalently, a = b, and c = 0.
9
5. If A, B, C are three dihedral angles of a trihedral angle, show that sin A, sin B, sin C satisfy the triangle inequality. Solution . Let a, b, c be the face angles of the trihedral angle opposite to A, B , C respectively. Since sin a sin b sin c = = sin A sin B sin C by the Law of Sines for spherical triangles, it suffices to show that sin b + sin c > sin a, or 1 1 1 1 2sin (b + c)cos (b c) > 2sin a cos a, 2 2 2 2 for any labelling of the angles. We now use the following properties of a, b, c:
−
(i) they satisfy the triangle inequality, (ii) 0 < a + b + c < 2π. Hence, cos 12 (b
− c) > cos 12 a. To complete the proof, we show that 1 1 sin (b + c) > sin a. 2 2
This follows immediately if b + c 1 sin (b + c) = sin π 2
−
≤ π; if b + c > π, then
1 (b + c) 2
1 > sin a 2
since π
−
b+c a > 2 2
.
Comment : More generally, if a1 , a2 , . . . , an are the sides of a spherical n-gon (convex), it then follows by induction over n that sin a1 + sin a2 +
·· · + sin an > 2sin ai,
i = 1, 2, . . . , n .
It also follows by induction that
| sin a1| + | sin a2| + · ·· + | sin an| > | sin(a1 + a2 + ·· · + an )| for any angles a 1 , a2 , . . . , an .
6. Are there any integral solutions (x,y,z) of the Diophantine equation (x
− y − z)3 = 27xyz
other than ( a,a,a) or such that xy z = 0?
−
Solution . Let x = u 3 , y = v 3 , z = w 3 , so that u 3
− v3 − w3 = 3uvw or equivalently (u − v − w)((u + v)2 + (u + w)2 + (v − w)2 ) = 0.
Hence an infinite class of non-trivial solutions is given by x = (v + w)3 ,
y = v 3 ,
z = w 3 .
Whether or not there are any other solutions is an open problem.
10
7. Does the Diophantine equation (x
− y − z)(x − y + z)(x + y − z) = 8xyz
have an infinite number of relatively prime solutions? Solution . By inspection, we have the trivial solutions (x,y,z) = ( 1, 1, 0)
± ±
and permutations thereof. equations
For other solutions, note that each of the three
− y − z = 2√ yz, x − y + z = 2√ xz, x + y − z = 2√ xy √ √ √ is satisfied by x = y + z. Consequently, we also have the infinite set of x
solutions y = m 2 ,
z = n 2 ,
x = (m + n)2
where (m, n) = 1.
It is an open problem whether or not there are any other infinite sets of relatively prime solutions.
8.
It is an easy result using calculus that if a polynomial P (x) is divisible by its derivative P (x), then P (x) must be of the form a(x r)n . Starting from the known result that P (x) 1 = P (x) x ri
−
−
where the sum is over all the zeros ri of P (x) counting multiplicities, give a non-calculus proof of the above result. Solution . Since P (x) is of degree one less than that of P (x), P (x) 1 = = P (x) a(x r)
−
1
x
Now letting x be the same.
→ any ri it follows that r = r i .
− ri
.
Hence all the zeros of P (x) must
9. Solve the simultaneous equations x2 (y + z) = 1,
y 2 (z + x) = 8,
z 2 (x + y) = 13.
Solution . More generally we can replace the constants 1, 8, 13 by a3 , b3 , c3 , respectively. Then by addition of the three equations and by multiplication of the three equations, we respectively get
x2 y = a 3 + b3 + c3 ,
(xyz)2 2xyz +
x2 y = (abc)3 ,
11
where the sums are symmetric over x, y , z . Hence, 2t3 + t2 (a3 + b3 + c3 ) = (abc)3
(1)
where t = xyz . In terms of t, the original equations can be rewritten as a3 tx
3 3 − y1 − z1 = 0, −x1 + tyb − 1z = 0, −x1 − y1 + ctz = 0.
These latter homogeneous equations are consistent since the eliminant is equation (1). Solving the last two equations for y and z , we get y =
x(b1 c1 1) , c1 + 1
−
x(b1 c1 2) b1 + 1
−
z =
where b 1 = b 3 /t, c 1 = c 3 /t. On substituting back in x 2 (y + z) = a 3 , we obtain x 3 and then x, y, z .
10. Determine the area of a triangle of sides a, b, c and semiperimeter s if (s
− b)(s − c) = a/h,
(s
− c)(s − a) = b/k,
(s
− a)(s − b) = c/l,
where h, k, l, are consistent given constants. Solution . h = k = l =
(s
−
a b)(s
1
(s
− c) 1
(s
− a)
− c)
+ +
1
=
(s
1
(s
− b)
+
1
(s
− c) ,
− a) , 1
(s
− b) .
Hence, h, k, l must satisfy the triangle inequality. Letting 2 s = h +k + l, it follows by addition that 1 1 1 s = + + (s a) (s b) (s c)
−
and then s
− a = (s −1 h) ,
s
−
−
− b = (s −1 k) ,
s
− c = (s 1− l) .
Adding the latter three equations, we get s =
1
1
1
+ + . (s − h) (s − k) (s − l)
Finally, the area of the triangle is given by
{ − a)(s − b)(s − c)}1/2 =
∆ = s(s
1
1
1
−h) + (s −k) + (s −l) (s − h)(s − k)(s − l)
(s
1/2
.
12
11. Prove that 3(x2 y + y 2 z + z 2 x)(xy 2 + yz 2 + zx2) where x, y,z
≥ xyz(x + y + z)2
≥ 0.
Solution . By Cauchy’s inequality (x2 y + y 2 z = z 2 x)(zx 2 + xy 2 + yz 2)
≥ (x2 √ yz + y2√ zx + z2√ xy)2 .
Hence it suffices to show that
(x3/2 + y 3/2 + z 3/2 3
2
≥
(x + y + z) 3
3
.
But this follows immediately from the power mean inequality. There is equality iff x = y = z.
12.
Determine all integral solutions of the Diophantine equation (x8 + y 8 + z 8 ) = 2(x16 + y 16 + z 16 ).
Solution . More generally one can find all integral solutions of (x2n + y 2n + z 2n )2 = 2(x4n + y 4n + z 4n ),
(2)
where n is a positive integer provided Fermat’s equation xn + y n = z n does not have any integer solutions for particular values of n > 2 chosen. Equation (2) can be rewritten as (xn + y n + z n )(y n + z n
− xn)(zn + xn − yn)(xn + yn − z n) = 0. The trivial solutions occur for (x,y,z) = (±a,a, 0) and permutations thereof.
(3)
For n = 1, any factor of the left hand side of (3) can be zero. For n = 2, (x,y,z) can be m2 + n2 ) in any order.
± the sides of any integral right triangle (2 mn, m 2 − n2,
Since Fermat’s equation is at least known not to have any non-trivial solutions for all n > 2 and < 100 and integral multiples thereof, there are not any non-trivial solutions for at least these cases.
13. Determine all the roots of the quintic equation 31x5 + 165x4 + 310x3 + 330x2 + 155x + 33 = 0. Solution . Since the equation can be rewritten as (x x 1 = 2ω r , x+1
−
− 1)5 = 32(x + 1)5,
r = 0, 1, 2, 3, 4
13
where ω is a primitive 5th root of unity. Hence, 1 + 2ω r x = , 1 2ω r
r = 0, 1, 2, 3, 4.
−
More generally the equation ax6 + 5bcx4 + 10ac2 x3 + 10bc3x2 + 5ac4 x + bc5 = 0 is the same as (b
− a)(x − c)5 = (b + a)(x + c)5.
14.
If F (x) and G(x) are polynomials with integer coefficients such that F (k)/G(k) is an integer for k = 1, 2, 3 . . . , prove that G(x) divides F (x). Solution . By taking k sufficiently large it follows that the degree of F is the degree of G. Then by the remainder theorem,
≥
F (x) Q(x) R(x) = + G(x) a G(x) where Q(x) is an integral polynomial, a is an integer, and R(x) is a polynomial whose degree is less than that of G(x). Now R(x) must identically vanish otherwise by taking k sufficiently large, we can make R(k)/G(k) arbitrarily small and this cannot add with Q(k)/a to be an integer.
15. Given that ABCDEF is a skew hexagon such that each pair of opposite sides are equal and parallel. Prove that the midpoints of the six sides are coplanar. Solution . Since each pair of opposite sides form a parallelogram whose diagonals bisect each other, all three different diagonals are concurrent say at point P . We now let A, B, C, A, B, C be vectors from P to A, B, C , D, E , F , respectively. The successive midpoints (multiplied by 2) are given by
− − −
A + B,
B + C,
C
− A, −A − B, −B − C, −C + A
and which incidentally form another centrosymmetric hexagon. It is enough now to note that (A + B) (B + C) + (C A) = 0.
16.
−
−
If a, b, c, d are the lengths of sides of a quadrilateral, show that
√ a √ , (4 + a)
√ b √ , (4 + b)
√ c √ , (4 + c)
√ d √ , (4 + d)
are possible lengths of sides of another quadrilateral. Solution . More generally one can show that if a 1 , a2 , . . . , an are the lengths of sides of an n-gon, then F (a1 ), F (a2 ), . . . , F ( an ) are possible lengths of sides of another n-gon where F (x) is an increasing concave function of x for x 0 and F (0) = 0. If a 1 is the largest of the a i ’s, then it suffices to show that F (a2 ) + F (a3 ) +
· ·· + F (an) ≥ F (a1).
≥
14
By the majorization inequality, we have F (a2 ) + F (a3 ) +
·· · + F (an) ≥ F (a2 + a3 + · · · + an) + (n − 2)F (0). Finally, F (a2 + a3 + · · · + an ) ≥ F (a1 ). Some admissable functions are F (x) = x α
and
xα k 2 + xα
for 0 < α < 1,
x , 1 (x + k 2 )
2
− e−k x, tanh x.
17.
Determine the maximum value of the sum of the cosines of the six dihedral angles of a tetrahedron. Solution . Let A, B, C, D be unit outward vectors normal to the faces of a tetrahedron ABCD. Then (xA + yB + zC + wD)2
≥ 0.
Expanding out and noting that A B = cos CD (here C D denotes the dihedral angle of which the side C D is an edge), etc., we get
·
x2 + y 2 + z 2 + w2
−
≥ 2xy cos CD + 2xz cos BD + 2xw cos BC
(1)
+2yz cos AD + 2yw cos AC + 2zw cos AB.
Setting x = y = z = w, we get that the sum of the cosines of the 6 dihedral angles is 2. There is equality iff A + B + C + D = 0. Since as known
≤
F a A + F b B + F c C + F d D = 0 where F a denotes the area of the face of the tetrahedron opposite A, etc., it follows that there is equality iff the four faces have equal area or that the tetrahedron is isosceles. Comment . In a similar fashion one can extend inequality (1) to n dimensions and then show that the sum of the cosines of the n(n + 1)/2 dihedral angles of an n-dimensional simplex is (n + 1)/2. Here the dihedral angles are the angles between pairs of (n 1)-dimensional faces and there is equality iff all the (n 1)-dimensional faces have the same volume.
≤
−
−
18. Which is larger
√ 3
( 2
− 1)1/3
or
− 3
1/9
3
2/9 +
3
Solution . That they are equal is an identity of Ramanujan. Letting x =
3
1/3 and y =
or equivalently that
3
2/3, it suffices to show that
√
− 1)1/3 = x3 + y3 = 1, √ √ ( 2 + 1)3 ( 2 − 1) = 3, 3
(x + y)( 2 3
3
4/9?
15
which follows by expanding out the left hand side. For other related radical identities of Ramanujan, see Susan Landau, How to tangle with a nested radical , Math. Intelligencer, 16 (1994), pp. 49–54.
19. Prove that 3min
a b c b c a + + , + + b c a a b c
≥
(a + b + c)
1 1 1 + + a b c
,
where a, b, c are sides of a triangle. Solution . Each of the inequalities 3 3
a b c + + b c a b c a + + a b c
≥ ≥
(a + b + c) (a + b + c)
1 1 1 + + a b c 1 1 1 + + a b c
, ,
follow from their equivalent forms (which follow by expansion):
− c)(c − a)2 + (c + b − a)(a − b)2 + (a + c − b)(b − c)2 ≥ 0, (b + c − a)(c − a)2 + (c + a − b)(a − b)2 + (a + b − c)(b − c)2 ≥ 0. (b + a
20. Let ω = e iπ/13 . Express 1−1 ω as a polynomial in ω with integral coefficients. Solution . We have
2 (1
−
(1 ω 13 ) = 1 + ω + ω 2 + ω) (1 ω) (1 + ω 13 ) 0= = 1 ω + ω 2 (1 + ω) =
− −
· · · + ω12,
− · · · + ω12.
−
Adding or subtracting, we get 1 (1
− ω)
= 1 + ω2 + ω4 + = ω + ω 3 +
·· · + ω12
· ·· + ω11.
More generally, if ω = e iπ/(2n+1) , 1 (1
− ω)
= 1 + ω2 + ω4 +
· ·· + ω2n.
21. x2
Determine all integral solutions of the simultaneous Diophantine equations + y 2 + z 2 = 2w2 and x 4 + y 4 + z 4 = 2w4 .
Solution . Eliminating w we get 2y 2 z 2 + 2z 2 x2 + 2x2 y 2
− x4 − y4 − z4 = 0
16
or (x + y + z)(y + z
− x)(z + x − y)(x + y − z) = 0,
so that in general we can take z = x + y. Note that if (x , y , z , w) is a solution, so is ( x, y, z, w) and permutations of the x, y , z . Substituting back, we get
± ± ± ±
x2 + xy + y 2 = w 2 . Since (x,y,w) = (1, 1, 1) is one solution, the general solution is obtained by the method of Desboves, that is, we set x = r + p, y = r + q and w = r. +q ) This gives r = ( p (+q pq − p) . On rationalizing the solutions (since the equation is homogeneous), we get
−
2
−
2
x = p2 + pq + q 2 + p(q p) = q 2 + 2 pq, y = p2 + pq + q 2 q (q p) = p 2 + 2 pq, w = p2 + pq + q 2 ,
−
− −
−
z = q 2
− p2.
22.
Prove that if the line joining the incentre to the centroid of a triangle is parallel to one of the sides of the triangle, then the sides are in arithmetic progression and, conversely, if the sides of a triangle are in arithmetic progression then the line joining the incentre to the centroid is parallel to one of the sides of the triangle. Solution . Let A, B, C denote vectors to the respective vertices A, B, C of the triangle from a point outside the plane of the triangle. Then the incentre I and the centroid G have the respective vector representations I and G, where I =
(aA + bB + cC) , (a + b + c)
G =
(where a, b, c are sides of the triangle). If G out (b + c
(A + B + C) , 3
− I = k(A − B), then by expanding
− 2a − k)A + (a + c − 2b + k)B + (a + b − 2c)C = 0,
where k = 3k(a + b + c). Since A, B, C are linearly independent, the coefficient of C must vanish so that the sides are in arithmetic progression. Also then k = b + c 2a = 2b a c.
−
− −
Conversely, if 2c = a + b, then G side AB.
− I = 3(A6(−a+Bb)(+bc−) a) , so that GI is parallel to the
17
23. Determine integral solutions of the Diophantine equation x y y z z w w x + + + =0 x + y y + z z + w w + x
−
−
−
−
(joint problem with Emeric Deutsch, Polytechnic University of Brooklyn). Solution . It follows by inspection that x = z and y = w are two solutions. To find the remaining solution(s), we multiply the given equation by the least common denominator to give P (x,y,z,w) = 0, where P is the 4th degree polynomial in x, y, z, w which is skew symmetric in x and z and also in y and w. Hence, P (x , y , z , w) = (x
− z)(y − w)Q(x , y , z , w),
where Q is a quadratic polynomial. On calculating the coefficient of x2 in P , we get 2z(y w). Similarly the coefficient of y 2 is 2w(x z), so that
−
− − P (x,y,z,w) = 2(x − z)(y − w)(xz − yw).
Hence, the third and remaining solution is given by xz = yw.
24. For x,y,z > 0, prove that 1 (i) 1 + (x + 1)
1 1+ x(x + 2)
≥
x
,
(ii) [(x + y)(x + z)]x [(y + z)(y + x)]y [(z + x)(z + y)]z
≥ [4xy]x[4yz]y [4zx]z .
Solution . Both inequalities will follow by a judicious application of the weighted arithmetic-geometric mean inequality (W–A.M.–G.M.) which for three weights is a b
u v w where a,b,c, u, v,w
c
≤
au + bv + cw a + b + c
a+b+c
,
≥ 0.
(i) The inequality can be rewritten in the more attractive form 1 1+ x
x
1 1+ x+1
≤
x+1
,
and which now follows by the W–A.M.–G.M. x+1
≤ 1 1+ x
x
1+x 1+ 1+x
1 x
1 = 1+ x+1
x+1
.
(ii) Also, the inequality here can be rewritten in the more attractive form 2x z + x
z +x
2y x+y
x+y
2z y + z
y +z
≤ 1.
18
But this follows by applying the W–A.M.–G.M. to 1=
[z + x]
2x z + x
[z + x].
25.
If ABCD is a quadrilateral inscribed in a circle, prove that the four lines joining each vertex to the nine point centre of the triangle formed by the other three vertices are concurrent. Solution . The given result still holds if we replace the nine point centres by either the orthocentres or the centroids. A vector representation is particularly `a propos here, since (with the circumcentre O as an origin and F denoting the vector from O to any point F ) the orthocentre H a , the nine point centre N a , the centroid Ga of BC D are given simply by Ha = B + C + D, N a = (B + C + D)/2, G a = (B + C + D)/3, respectively, and similarly for the other three triangles. Since the proofs for each of the three cases are practically identical, we just give the one for the orthocentres. The vector equation of the line L a joining A to H a is given by L a = A + λa [B + C + D A] where λa is a real parameter. By letting λa = 1/2, one point on the line is [A + B + C + D]/2 and similarly this point is on the other three lines. For the nine point centres, the point of concurrency will be 2[ A + B + C + D]/3, while for the centroids, the point of concurrency will be 3[ A + B + C + D]/4.
−
26. How many six digit perfect squares are there each having the property that if each digit is increased by one, the resulting number is also a perfect square? Solution . If the six digit square is given by m2 = a 105 + b 104 + c 103 + d 102 + e 10 + f,
·
·
·
·
·
then n2 = (a + 1) 105 + (b + 1) 104 + (c + 1) 103 + (d + 1) 102 + (e + 1) 1 0 + (d + 1),
·
·
·
·
·
so that n2
− m2 = 111, 111 = (111)(1, 001) = (3 · 37)(7 · 11 · 13).
Hence, n + m = d i
and n
− m = 111, 111/di
where di is one of the divisors of 111, 111. Since 111, 111 is a product of five primes it has 32 different divisors. But since we must have d i > 111, 111/di, there are at most 16 solutions given by the form lm = 12 (di 111, 111/di). Then since m 2 is a six digit number, we must have
−
632.46
≈ 200
√
10 < 2m < 2, 000.
On checking the various divisors, there are four solutions. One of them corresponds to d i = 3 13 37 = 1, 443 so that m = 12 (1, 443 7 11) = 683 and m 2 = 466, 489.
· ·
− ·
19
Then, 466, 489 + 111, 111 = 577, 600 = 760 2. The others are given by the table m m2 317 100, 489 565 319, 225 445 198, 025
di 3 7 37 = 777 3 11 37 = 1, 221 7 11 13 = 1, 001
· · · · · ·
n2 n 211, 600 460 430, 336 656 309, 136 556
27.
Let vi wi , i = 1, 2, 3, 4, denote four cevians of a tetrahedron v1 v2 v3 v4 which are concurrent at an interior point P of the tetrahedron. Prove that pw1 + pw2 + pw3 + pw4
≤ max viwi ≤ longest edge.
Solution . We choose an origin, o, outside of the space of the tetrahedron and use the set of 4 linearly independent vectors Vi = ovi as a basis. Also the vector from o to any point q will be denoted by Q. The interior point p is then given by P = x 1 V1 + x2 V2 + x3 V3 + x4 V4 where xi > 0 and i xi = 1. It now follows that x V Wi = P− (for other properties of concurrent cevians via vectors, see [1987: 1−x 274–275]) and then that i
i
i
pwi =
P xi Vi 1 xi
vi wi = Summing
− −
−P
P xi Vi 1 xi
− −
=
− V i
≤ pwi =
i
xi
i
xj
j
xi (P Vi ) = xi 1 xi
− −
=
Vj 1
P 1
− Vi − xi
− Vi − xi
=
=
i
xj
j
xj
j
xi (vi wi )
Vj 1
− Vi − xi
Vj 1
− Vi − xi
,
.
vi wi , ≤ max i
and with equality only if v i wi is constant. Also, xj
vi wi
1
j =i
− xi
max Vr r
| − Vi | = max |V r − V i |. r
Finally,
pwi
≤ max vi wi ≤ max |Vr − Vs |. r,s i
Comment: In a similar fashion, it can be shown that the result generalizes to n-dimensional simplexes. The results for triangles are due to Paul Erd˝os, Amer. Math. Monthly, Problem 3746, 1937, p. 400; Problem 3848, 1940, p. 575.
20
28.
Determine the radius r of a circle inscribed in a given quadrilateral if the lengths of successive tangents from the vertices of the quadrilateral to the circle are a,a,b,b,c,c,d,d, respectively. Solution . Let 2A, 2B, 2C , 2D denote the angles between successive pairs of radii vectors to the points of tangency and let r be the inradius. Then r =
a b c d = = = . tan A tan B tan C tan D
Also, since A + B + C + D = π, we have tan(A + B) = tan(C + D) = 0, or tan A + tan B tan C + tan D + = 0, 1 tan A tan B 1 tan C tan D
−
−
so that r(a + b) r(c + d) + 2 = 0. r2 ab r cd
−
−
Finally, r2 =
abc + bcd + cda + dab . a + b + c + d
29. Determine the four roots of the equation x 4 + 16x − 12 = 0. Solution . Since x4 + 16x
− 12 = (x2 + 2)2 − 4(x − 2)2 = (x2 + 2x − 2)(x2 − 2x + 6) = 0, √ √ the four roots are −1 ± 3 and 1 ± i 5. 30.
Prove that the smallest regular n-gon which can be inscribed in a given regular n-gon is one whose vertices are the midpoints of the sides of the given regular n-gon. Solution . The circumcircle of the inscribed regular n-gon must intersect each side of the given regular n-gon. The smallest that such a circle can be is the inscribed circle of the given n-gon, and it touches each of its sides at its midpoints.
31. If 311995 divides a 2 + b2, prove that 311996 divides ab. Solution . If one calculates 1 2 , 22 , . . . , 302 mod 31 one finds that the sum of no two of these equals 0 mod 31. Hence, a = 31a1 and b = 31b1 so that 311993 divides a21 + b21 . Then, a 1 = 31a2 and b 1 = 31b2 . Continuing in this fashion (with p = 31), we must have a = p 998m and b = p 998n so that ab is divisible by p 1996 . More generally, if a prime p = 4k + 3 divides a 2 + b2 , then both a and b must be divisible by p. This follows from the result that “a natural n is the sum of squares of two relatively prime natural numbers if and only if n is divisible neither by 4 nor by a natural number of the form 4k + 3” (see J.W. Sierpi´ nski, Elementary Theory of Numbers, Hafner, NY, 1964, p. 170).
21
32. Determine the minimum value of S =
( + 1) + 2( − 2) + ( + 3) + ( + 1) + 2( − 2) + ( + 3) ) + ( + 1) + 2( − 2) + ( + 3) + ( + 1) + 2( − 2) + ( + 3) 2
a
2
c
2
b
2
c
2
d
2
a
2
b
2
c
2
d
2
a
2
d
b
2
where a, b, c, d are any real numbers. Solution . Applying Minkowski’s inequality, S
≥
(4 + s)2 + 2(s
− 8)2 + (s + 12)2 =
4s2 + 288
√
where s = a + b + c + d. Consequently, min S = 12 2 and is taken on for a = b = c = d = 0.
33. A set of 500 real numbers is such that any number in the set is greater than one-fifth the sum of all the other numbers in the set. Determine the least number of negative numbers in the set. Solution . Letting a 1 , a2 , a3 , . . . denote the numbers of the set and S the sum of all the numbers in the set, we have a1 >
S
− a1 ,
a2 >
5
S
− a 2 , 5
...,
a6 >
S
− a6 . 5
Adding, we get 0 > S a1 a2 a6 so that if there were six or less negative numbers in the set, the right hand side of the inequality could be positive. Hence, there must be at least seven negative numbers.
− − −···−
Comment. This problem where the “5” is replaced by “1” is due to Mark Kantrowitz, Carnegie–Mellon University.
34. Prove that a+b+c
≥
b2 + c2
− a2 +
− b2 +
− b2 ≤
2c ,
c2 + a2
where a, b, c are sides of a non-obtuse triangle.
a2 + b2
− c2 ,
Solution . By the power mean inequality
b 2 + c2
− a2 +
c2 + a2
and similarly two other such inequalities. Then, adding, we get the desired result.
35. Determine the extreme values of the area of a triangle ABC , given the lengths of the two altitudes h c , h b . [Correction to question made.] Solution . Let D and E be the feet of the altitudes hb and hc . Then by the Law h h of Sines applied to triangles ABD and ACE , c = sin A and b = sin A . Twice the h h h h area is given by 2[ABC ] = sin occuring for A . Hence, the minimum area is 2 π A = 2 . Also, by letting A approach π, the area becomes unbounded. In this case a would be arbitrarily large. b
b
c
c
b
c
22
36.
Determine the maximum area of a triangle AB C given the perimeter p and the angle A. Solution . Since 2[ABC ] = bc sin A, we have to maximize bc subject to p = a + b + c,
a 2 = b 2 + c2
and
≥ √
− 2bc cos A .
Since p a = b + c 2 bc, bc will be a maximum when b = c regardless of the value of a. Thus, we have
−
p = a + 2b Then ( p
a2 = 2b2
and
− 2b2 cos A .
− 2b)2 = 2b2 − 2b2 cos A. Solving for b: b =
p 1 + sin A2
2cos2
A 2
so that p2 tan A2
1 + sin A2
max[ABC ] =
4cos2
A 2
2
.
37. Determine the minimum value of
(a2 + a3 + a4 + a5 ) a1
1/2
where the sum is cyclic over the positive numbers a 1 , a2 , a3 , a 4 , a 5 . Solution . Applying the AM–GM Inequality to each term of the sum, the given sum is greater than or equal to
2
a2 a3 a4 a5 a41
1/8
(where again the sum is cyclic). Finally applying the AM–GM Inequality again, the latter sum is greater than or equal to 10. There is equality in the given inequality if and only if the ai ’s are equal. In a similar fashion it follows that if we increase the number of variables to n + 1 and change the 1/2 power to any positive number p, the minimum here would be (n + 1) p .
38. ABCD and AB C D are any two parallelograms in a plane with A opposite to C and C . Prove that B B , C C and DD are possible sides of a triangle.
Solution . Let the vectors from A to B and A to D be denoted by U 1 and U 2 , and the vectors from A to B and A to D be denoted by V 1 and V 2 . Then BB = V1
− U1, DD
= V2
− U2 ,
and CC = V1 + V2
− U1 − U2 , so that CC = BB +DD . The rest follows from the triangle inequality |P ± Q| ≤ |P| + |Q| and with equality only if P and Q have the same direction.
23
Remarks : Christopher J. Bradley of Clifton College, Bristol, UK, also submitted solutions for the quickies, noting a problem with problem 2. He points out that he gave problem 1 to a group of students in 1988.
39. Determine the maximum value of S = 4(a4 + b4 + c4 + d4 )
− (a2 bc + b2cd + c2da + d2ab) − (a2b + b2c + c2d + d2 a) ,
where 1
≥ a, b, c, d ≥ 0. Solution . S ≤ 4(a2 + b2 + c2 + d2 ) − (a2 b2 c2 + b2 c2 d2 + c2 d2 a2 + d2 a2 b2 ) −(a2b2 + b2c2 + c2 d2 + d2a2).
Since the expression on the right hand side is linear in a2 , b2 , c2 , and d2 , it takes on its maximum at the endpoints 0, 1 for each variable. By inspection, S max = 9 and is taken on for a = b = c = 1 and d = 0.
40.
If a, b, c, d are > 0, prove or disprove the two inequalities:
ab bc cd da c + d + a + b (ii) a 2 b + b2 c + c2 d +
(i)
≥
d2 a
a + b + c + d,
≥
abc + bcd + cda + dab.
Solution . Neither inequality is valid. (i) Just consider the case: b = 2, c = 5, d = 1 and a is very large. (ii) Just consider the case: a = 2, b = 1, c = 8 and d is very small.
41.
Determine all the points P (x,y,z), if any, such that all the points of tangency of the enveloping (tangent) cone from P to the ellipsoid y x z a + b + c = 1 (a > b > c), are coplanar. 2
2
2
2
2
2
Solution . Consider the affine transformation x = xa , y = yb , z = zc which takes the ellipsoid into a sphere. Under this transformation, lines go to lines, planes go to planes, and tangency is preserved. Consequently, any enveloping cone of the ellipsoid goes into an enveloping cone of the sphere and which by symmetry is a right circular one and its points of tangency are a circle (coplanar) of the sphere. Thus, P can be any exterior point of the ellipsoid.
42. Determine whether or not there exists a set of 777 distinct positive integers such that for every seven of them, their product is divisible by their sum. Solution . Just take any 777 distinct positive integers and multiply each one by the product of the sums of every 7 of them.
43.
If R is any non-negative rational approximation to better rational approximation.
√ 5, determine an always
Solution . Assuming the better approximation has the form are rational, we must satisfy
→ √
aR + b cR + d
−
√
aR+b where cR+d
a, b, c, d
√ | − 5| .
5 < R
(2)
√
√
5+b If R 5, the left hand side must 0. Thus, we must have ac√ 5+ = 5, so d that d = a and b = 5c. Then substituting these values in (2) and dividing both
→
24
| − √ 5|, we get √ |cR + a| > |c 5 − a|
sides by the common factor R
and which can easily be satisfied by letting a = 2 and c = 1. Finally, our better +5 approximation is 2RR+2 .
44.
A sphere of radius R is tangent to each of three concurrent mutually orthogonal lines. Determine the distance D between the point of concurrence and the centre of the sphere. Solution . Let the three lines be the x, y, and z–axes of a rectilinear coordinate system and the equation of the sphere be (x a)2 + (y b)2 + (z c)2 = R 2 . The required distance squared is a 2 + b2 + c2 . Since the distance from the centre of the sphere to each of the lines is R, we have
−
−
−
R2 = b2 + c2 = c2 + a2 = a2 + b2 . Hence, D2 = 3R2/2.
45.If P (x , y , z , t) is a polynomial in x, y, z, t such that P (x , y , z , t) = 0 for all real x, y , z, t satisfying x 2 + y 2 + z 2
− t2 = 0, prove that P (x , y , z , t) is divisible by x2 + y 2 + z 2 − t2 .
Solution . By the Remainder Theorem, P (x , y , z , t) = (x2 + y 2 + z 2
− t2)Q(x , y , z , t) + R(x,y,z)t + S (x,y,z)
where Q, R and S are polynomials. Now, letting t be successively /2 2 2 2 1 (x + y + z ) , it follows that R = S = 0.
±
46. From a variable point P on a diameter AB of a given circle of radius r, two segments P Q and P R are drawn terminating on the circle such that the angles QP A and RP B are equal to a given angle θ. Determine the maximum length of the chord QR. Solution . Extend the chords QP and RP to intersect the circle again at points Q and R . It now follows that the arcs QR and Q R are congruent and thus, their measures are π 2θ. Then if O is the centre, triangle OQR is isosceles whose vertex angle is also π 2θ. Hence, QR = 2r cos θ, which is the same for all P . (rs)! is an integer, where r, s are positive integers, prove that 47. Using that s!(r!) s (rst)! is an integer for positive integers r, s, t. t!(s!)t (r!)ts Solution . It follows from the given relation that both
−
−
(r(st))! (st)!(r!)st
and
are integers. Now just multiply them together.
(st)! t!(s!)t
25
y) 48. Determine the range of tan(x + tan x sin y =
given that
√ 2 sin(2x +
y) .
Solution . Since sin y sin(2x + y)
sin(x + y)cos x cos(x + y)sin x sin(x + y)cos x + cos(x + y)sin x tan(x + y) tan x = = 2 , tan(x + y) + tan x
−
=
√
−
so that
tan(x + y) tan x
=
−3 − 2
√
2 = a constant.
49.
A, B , C are acute angles such that sin 2 A + sin2 B + sin2 C = 2. Prove that A + B + C < 180 ◦ . Solution . Equivalently, cos2 A + cos2 B + cos2 C = 1. Then, using the identity for arbitrary angles A, B , C , cos2 A + cos2 B + cos2 C 1 + 2 cos A cos B cos C = cos(S ) cos(S A) cos(S B)cos(S C ) ,
−
−
−
−
where 2S = A + B + C , we have cos A cos B cos C = 2cos(S ) cos(S
− A)cos(S − B) cos(S − C ) .
Since the left-hand-side is positive, it follows that 2 S < π.
50. Determine the maximum and minimum values of
a2 cos2 θ + b2 sin2 θ +
where a and b are given constants. Solution . Let S = get, 2
2
a2 sin2 θ + b2 cos2 θ
a2 cos2 θ + b2 sin2 θ +
2
S = a + b + 2 Since cos4 θ + sin4 θ = 1 written as
a2 sin2 θ + b2 cos2 θ. Squaring we
(a4 + b4 )sin2 θ cos2 θ + a2 b2 (cos4 θ + sin4 θ) .
− 2sin2 θ cos2 θ, the expression inside the radical can be (a2 − b2 )2 (sin2 2θ) + a2 b2 .
4 Hence, the maximum and minimum are taken on for = giving S max = 2(a2 + b2 ) and S min = a + b.
x 51. Determine the range of values of tan3 tan x
π 4 and
0, respectively,
for x in [0, 2π].
Solution . Expanding out: (tan3x) = tan x
Since tan x lies in the interval [0,
1+
8 (1 3 tan2 x)
−
3
.
∞), the range consists of all values ≤ 13 and ≥ 3.
26
52. How large can the sum of the angles of a spherical right triangle be? Solution . Consider a lune with angle π2 . Now draw a very small arc across it near one of its vertices producing a right triangle of area nearly equal to π (for a unit sphere). Hence the spherical excess A + B + C π = π ε. Hence the sum must be less than 2π.
−
−
53.
Let ABC be a spherical triangle whose mid-points of the sides are A , B , and C . If B C is a quadrant π2 , find the maximum value of A B + A C .
−→ −→ −→
Solution . Let A , B , C be unit vectors from the centre of the sphere to the
−A→ + −B→ −→ −→ vertices A, B, C of the spherical triangle. Then C and B are given by
−→ + −→ A
and
−A→ + −C → −A→ + −C → .
We now have that cos B C = 0 =
B
−→ −→ · −→ −→ A +B
A + C
=
1+cos a +cos b +cos c where a, b, c are the sides of ABC . Since this is symmetric π in A, B , C , A C = A B = , so that A C + A B = the constant π. 2
54. Let ABC be a triangle with centroid G. Determine the point P in the plane of ABC such that AP AG + BP BG + CP CG is a minimum and express this minimum value in terms of the side lengths of ABC .
· · · −→ −→ −→ −→ Solution . Let A , B , C , and P be vectors from G to A, B, C , and P , respectively. Then AP AG + BP BG + CP CG
· −→ −→ · −→ −→ · −→ −→ −→ −→ −→ = A A − P + B B − P + C C − P ≥ −A→ · −A→ − −P → + −B→ · −B→ − −P → + −C → · −C → − −P → −A→ 2 + −B→ 2 + −C → 2 − −P → · −A→ + −B→ + −C → = −A→ 2 + −B→ 2 + −C → 2 since −A→ + −B→ + −C → =
4(m2 + m2 + m2 )
−→
= 0 .
(a2 + b2 + c2 )
a c b Hence the minimum is = 9 3 attained when P coincides with G (ma is the median from A, etc.).
,
and
is
Comment . This was a short-listed problem for the 2001 IMO. It was eliminated after I gave the above Quickie solution.
28
1456. [1989 : 178] 1. Find a pair of integers (a, b) such that x13
− 233x − 144
and x15 + ax + b
have a common (non-constant) polynomial factor. 2. Is the solution unique? I. Solution to (a) by Mathew Englander, Kitchener, Ontario. [1990 : 250] Consider the Fibonacci sequence f 0 = f 1 = 1, f n+2 = f n+1 + f n . For n 0 let p n be the polynomial
≥
xn + xn−1 + 2xn−2 + 3xn−3 + 5xn−4 +
· · · + f n−1x + f n.
Then (x2
= xn+2 + xn+1 + 2xn + + f n−1 x3 + f n x2 (xn+1 + xn + 2xn−1 + + f n−1 x2 + f n x) (xn + xn−1 + 2xn−2 + + f n−1 x + f n )
− x − 1) pn
···
− −
= xn+2
· ·· · ··
− f n+1x − f n.
Observe that f 11 = 144, f 12 = 233, f 13 = 377, f 14 = 610. Thus, (x2
− x − 1) p11
= x13
− 233x − 144
(x2
− x − 1) p13
= x15
− 610x − 377.
and Thus, (a, b) = ( 610, 377) is one solution to the problem.
−
−
II. Solution to (b) by Stanley Rabinowitz, Westford, Massachusetts, USA. [1990 : 251] [Of course, Rabinowitz first answered part (a). —Ed.] My computer tells me that x 13
x2
−x−1 ×
− 233x − 144 factors as
x11 + x10 + 2x9 + 3x8 + 5x7 + 8x6 +13x5 + 21x4 + 34x3 + 55x2 + 89x + 144 , (1)
where both factors are irreducible. (This factorization, involving Fibonacci numbers, can be found in Charles R. Wall, Problem B–55, Fibonacci Quarterly 3 (1965) 158, and Russel Euler, Problem 8, Missouri Journal of Math. Sciences 1 (1989) 44–45.) If x 15 + ax + b has a (non-constant) polynomial factor in common with x 13 233x 144, then this common factor must be one of the two factors in (1) (or their product).
−
−
29
Case (i). Suppose that x15 + ax + b is divisible by x2 x 1. Then since x15 610x 377 is also divisible by x 2 x 1, the difference between these two fifteenth degree polynomials must also be divisible by x 2 x 1. In other words, (a + 610)x + (b + 377) would be divisible by x 2 x 1. The only way this could happen is if a = 610 and b = 377.
−
−
− −
−
− −
−
− − − −
Suppose that x 15 + ax + b is divisible by
Case (ii).
x11 + x10 + 2x9 + 3x8 + 5x7 + 8x6 + 13x5 + 21x4 + 34x3 + 55x2 + 89x + 144. In this case we have x15 + ax + b = (x4 + px3 + qx2 + rx + s) or
13 − 144 , · x x−2 233x −x−1
(x15 + ax + b)(x2
− x − 1 ) = (x4 + px3 + qx2 + rx + s)(x13 − 233x − 144). But this equation cannot hold, since the coefficient of x 5 is 0 on the left and −233 on the right.
Thus, the solution in part (a) is unique. Editor’s comment . [1990 : 252] In addition Richard I. Hess, Rancho Palos Verdes, California, USA, and Robert E. Shafer, Berkeley, California gave almost–complete proofs of part (b). Hess’s dealt with the stronger problem of showing that no integers (a, b) = ( 610, 377) exist such that x15 +ax+b has a root in common with x13 233x 144. Shafer’s solution was like Rabinowitz’s but without his computer, and thus, was much longer. Shafer also wonders about the irreducibility of the polynomial
−
−
−
xn
− F n+1x − F n x2 − x − 1
= xn−2 + xn−3 + 2xn−4 +
−
· · · + F n−1 x + F n,
where F n is the nth Fibonacci number. A computer verifies that these are all irreducible for n 13. Can anyone come up with further information?
≤
2054. [1995 : 202] Are there any integral solutions of the Diophantine equation (x + y + z)3 = 9 x2 y + y 2 z + z 2 x
other than (x,y,z) = (n,n,n)?
I. Solution by Adrian Chan, student, Upper Canada College, Toronto, Ontario. [1996 : 188] No, there are no integral solutions other than (x,y,z) = (n,n,n). Without loss of generality, let x y and x z. Let y = x + a and z = x + b, where a and b are non-negative integers. Then the given equation becomes
≤
≤
(3x + a + b)3 = 9 x2 (x + a) + (x + a)2 (x + b) + (x + b)2 x .
30
After expanding and simplifying, this is (a + b)3 = 9a2 b, or a3
− 6a2b + 3ab2 + b3
= 0.
(1)
Let a = kb, where k is a rational number. Then (1) becomes b3 (k3
− 6k2 + 3k + 1) = 0. By the Rational Roots Theorem, k 3 − 6k 2 + 3k + 1 = 0 does not have any rational roots. Thus, since k is rational, k 3 − 6k 2 + 3k + 1 = 0. Therefore, b = 0 and a = 0, so that x = y = z.
II. Solution by Murray S. Klamkin, University of Alberta, Edmonton, Alberta . [1996 : 188] Letting y = x + u and z = x + v, the equation reduces to (u + v)3 = 9u2 v,
(2)
where u and v are integers. We now show that the only solution to (2) is u = v = 0 so that (x,y,z) = (n,n,n) is the only solution of the given equation. Letting u + v = w, (2) becomes w3 = 9u2 (w u). (3)
−
Hence, w = 3w1 where w 1 is an integer, and (3) is 3w13 = u 2 (3w1 that u = 3u1 for some integer u 1 , and we get w13 = 9u21(w1
− u). It follows
− u1).
Comparing this equation to (3), we see by infinite descent that the only solution to (3) is u = w = 0, which gives the negative result.
1863. [1993 : 203] Are there any integer solutions of the equation (x + y + z)5 = 80xyz(x2 + y 2 + z 2 ) such that none of x, y , z are 0? Solution by Murray S. Klamkin, University of Alberta, Edmonton, Alberta. [1994 : 173] No. The identity (x + y + z)5
− (−x + y + z)5 − (x − y + z)5 − (x + y − z)5
= 80xyz(x2 + y 2 + z 2)
follows by expanding out. Hence, the given equation is equivalent to ( x + y + z)5 + (x
−
− y + z)5 + (x + y − z)5
= 0.
(1)
Since Fermat’s Last “Theorem” is known to be valid for exponent 5, there are only the trivial solutions x = 0, y + z = 0 and symmetrically. [For if x + y + z = 0, (1) becomes 0 5 + (2z)5 + (2y)5 = 0, which implies z = y, so that x = 0 = y + z.]
−
−
31
The mysterious procedure by which the proposer conjures up identities such as the one above seems to be unavailable to the rest of us! Just two readers guessed that the answer to the problem was no, and neither had a complete proof. P. Penning, Delft, the Netherlands, investigated when the number 80 of the problem can be replaced by a positive integer N so that the resulting equation has an all non-zero integer solution. Via a computer search, he found only two such values of N : N = 81, with solutions x = y = z; and N = 108, with solutions 4x = 4y = z. Any others ?
1027. [1985 : 83, 248] Determine all quadruples (a,b,c,d) of non-zero integers satisfying the Diophantine equation 1 1 1 1 abcd + + + a b c d
2
= (a + b + c + d)2
and such that a 2 + b2 + c2 + d2 is a prime. Solution par C. Festraets–Hamoir, Bruxelles, Belgique . [1986: 160] 2
1 1 1 1 = (a + b + c + d)2 abcd + + + a b c d 1 1 1 1 2 2 2 2 2 2 = abcd 2 + 2 + 2 + 2 + + + + + + a b c d ab ac ad bc bd cd
⇐⇒ ⇐=⇒ ⇐=⇒
= a2 + b2 + c2 + d2 + 2ab + 2ac + 2ad + 2bc + 2bd + 2cd bcd acd abd abc + + + = a2 + b2 + c2 + d2 a b c d b2 c2 d2 + a2 c2 d2 + a2 b2 d2 + a2 b2 c2 = abcd(a2 + b2 + c2 + d2 ).
Posons a 2 + b2 + c2 + d2 = p ( p premier). On a b2 c2 d2 + a2 c2 d2 + a2 b2 d2 + a2 b2 c2 0 (mod p) 2 2 2 2 2 2 2 2 b c (d + a ) + a d (c + b ) 0 (mod p) 2 2 2 2 2 2 2 2 b c ( b c ) + a d (c + b ) 0 (mod p) 2 2 2 2 2 2 (a d 0 (mod p) b c )(b + c ) 2 2 2 2 a d b c 0 (mod p) car (b2 + c2 ,p) = 1.
− − − − ≡
≡
≡ ≡
≡
De mˆ eme, on a a2 c2
− b2d2 ≡
0 (mod p).
D’o` u, par addition a2 d2 b2 c2 + a2 c2 b2 d2 0 (mod p) 2 2 2 2 (a b )(d + c ) 0 (mod p) a2 b2 0 (mod p) car (d2 + c2 ,p) = 1.
− − − ≡
− ≡
≡
32
Par sym´etrie, om obtient a2 b2 c2 d2 (mod p) p = a2 + b2 + c2 + d2 4a2 (mod p), a2 = 0 ou a 2 = p
≡
≡
≡
≡
ce qui est impossible. Donc, il n’existe aucun quadruple d’entiers positifs non nuls (a,b,c,d) satisfaisant les conditions donn´ees.
969. [1984 : 217] Find a 3-parameter solution of the Diophantine equation x y z + 2 + 2 = 2 2 2 x +w y +w z + w2
2w2 . (x2 + w2 )(y 2 + w2 )(z 2 + w2 )
(1)
Solution by Murray S. Klamkin, University of Alberta, Edmonton, Alberta (revised by the editor ) [1985 : 300] It is clear that (x , y , z , w) is a solution of (1) if and only if (kx,ky,kz,kw) is a solution for any k > 0, so that it will suffice for our problem to find a 3–parameter family of primitive solutions, corresponding to k = 1. We first show that (1) is satisfied by infinitely many solutions of yz + zx + xy = w2 ,
(2)
so that it will make sense to look for solutions of (1) among those of (2). It is clear that (1) is satisfied by any solution of (2) in which w = 0 and xy z = 0. For solutions of (2) with w = 0, we can choose x, y, z , such that
x = cot A, w
y = cot B, w
z = cot C, w
(3)
where A, B, C , are angles of a triangle. This can be done in infinitely many ways for each w = 0. Substituting (3) into (2) and then into (1) gives
cot B cot C + cot C cot A + cot A cot B = 1 and sin2A + sin 2B + sin 2C = 4sin A sin B sin C, respectively, both of which are valid triangle identities. Hence, every solution (3) satisfies (2) and (1). To solve (2), we use the method of Desboves and assume that x = a + λp,
y = a + λq,
z = λr,
w = a + λt,
where λ = 0, is a solution of (2). Substituting these values into (2) gives
λ =
a(2t p q 2r) . qr + rp + pq t2
− − − −
(4)
33
Since (2) is homogeneous, we can multiply (4) by qr + rp + pq
− t2 and obtain
− t2) + ap(2t − p − q − 2r), − t2) + aq (2t − p − q − 2r), z = ar(2t − p − q − 2r), 2 w = a(qr + rp + pq − t ) + at(2t − p − q − 2r). With m = t − p and n = t − q , we get x = ar(m − n) − m2 , y = ar(n − m) − n2 , z = ar(m + n − 2r), w = amn − r(m + n). x = a(qr + rp + pq y = a(qr + rp + pq
It will be found that these values satisfy (2) for all a, m, n, r, but they satisfy (1) for all m, n, r, if and only if a < 0. [To facilitate the latter verification process, we note that x2 + w2 y 2 + w2
= =
a2 (m2 + n2 )(m2 2rm + 2r2 ), a2 (m2 + n2 )(n2 2rn + 2r2 ),
z 2 + w2
=
a2 (m2
Consequently, we take a = solution set of (1):
− −
− 2rm + 2r2)(n2 − 2rn + 2r2)].
− 1 and obtain the following 3–parameter primitive x y z w
= m2 r(m n), = n2 r(n m), = r(2r m n), = r(m + n) mn.
− − − − − − −
Editor’s comment [1985 : 301] Equation (2) is extensively discussed in L.J. Mordell, Diophantine Equations , Academic Press, New York, 1969, pp. 291–292.
1561. [1990 : 204] Determine an infinite class of integer triples (x,y,z) satisfying the Diophantine equation x2 + y 2 + z 2 = 2yz + 2zx + 2xy 3.
−
Solution by Hayo Ahlburg, Benidorm, Spain. [1991 : 252] The well-publicized identity (see L´eo Sauv´e’s footnote on [1976 : 176]) 12 + (n2 n + 1)2 + (n2 + n + 1) 2 = 2(n2 n + 1) + 2(n2 n + 1)(n2 + n + 1)
−
−
2
−
+2(n + n + 1)
−3
(1)
34
is one answer to this problem if we choose n to be any integer. To find all solutions, we rearrange the original equation and get z2
− 2(x + y)z + (x − y)2 + 3
= 0
and z = x+y
±
4xy
− 3.
To make z an integer, 4xy 3 must be the square of an odd number 2 n + 1; that is, 4n2 + 4n + 1 = 4xy 3 or
−
−
n2 + n + 1 = xy, where n can be any integer. We can choose x and y as factors of n 2 + n + 1 (this can sometimes be done in several ways), and with z = x+y
± (2n + 1)
this solves the original equation. Due to the symmetry of the problem, any permutation of the values for x, y and z is also a solution. n, x, y and z can of course also have negative values. Using both signs for x, y,z, and for 2n +1 in the expression for z leads to duplications. But using + signs throughout already gives an infinite number of solutions. Factoring n 2 + n+ 1 into x = 1 and y = n 2 + n + 1, and with z = x + y (2n + 1), we get equation (1).
−
A nice special group is the series 1, 1, 3, 7, 19, 49, 129, 337, 883, 2311, 6051, . . ., where each term has the form F 2n+1 F n F n+1 made up of Fibonacci numbers. Any three consecutive numbers of this series form a solution.
−
2034.
[1995 : 130, 157] Murray S. Klamkin and M.V. Subbarao, University of Alberta, Edmonton, Alberta . (a) Find all sequences p 1 < p2 < < p n of distinct prime numbers such that
···
1 p1
1 p2
1 pn
· · · 1+
1+
1+
is an integer. (b) Can 1 1+ 2 a1
1 1+ 2 a2
· · ·
1 1+ 2 an
be an integer, where a 1 , a2 , . . . are distinct integers greater than 1? Solution to (a), by Heinz-J¨ urgen Seiffert, Berlin, Germany . [1996 : 138] If the considered product is an integer, then pn ( pi + 1) for some i 1, 2, . . ., n 1 . Since pi + 1 pn , it then follows that pn = pi + 1, which implies pi = 2 and pn = 3. Thus, n = 2, p1 = 2, p2 = 3. This is indeed a solution since 1 + 12 1 + 13 = 2.
− }
≤
|
∈ {
35
Solution to (b). [1996 : 138] In the following, we present four different solutions submitted by eight solvers and the proposers. In all of them, p denotes the given product, and it is shown that 1 < p < 2 and thus, p cannot be an integer. Clearly one may assume, without loss of generality, that 1 < a1 < a2 < .. . < an . Solution I, by Carl Bosley, student, Washburn Rural High School, Topeka, KS, USA; Kee-Wai Lau, Hong Kong; and Kathleen E. Lewis, SUNY, Oswego, New York. [1996 : 138] Since 1 + x < e x for x > 0, we have
−≈
1 1 1 + + . . . + a21 a22 a2n 1 1 < exp 2 + 2 + . . . 2 3 π2 = exp 1 1.90586 < 2. 6
1 < p < exp
Solution II, by Toby Gee, student, the John of Gaunt School, Trowbridge, England. [1996 : 138] Since a 1 2, we have
≥
n+1
1
n+1
1 1+ 2 k
≤ −
< p
k=2
n+1
=
k=2
k k + 1
=
k=2
n+1 k=2
k
k
−1
k2 + 1 < k2 =
n+1
k=2
k2 k2
−1
2(n + 1) < 2. n + 2
Solution III, by Walther Janous, Ursulinengymnasium, Innsbruck, Austria; V´aclav Kone¸ny ´ , Ferris State University, Big Rapids, MI, USA; Heinz-J¨ urgen Seiffert, Berlin, Germany; and Murray S. Klamkin and M.V. Subbarao, University of Alberta, Edmonton, Alberta. [1996 : 138] 1 < p <
∞
1 t2
∞
1 1 1+ 2 2 t=1 t
1+
t=2
=
=
sinh π 2π
≈ 1.838 < 2.
Solution IV, Richard I. Hess, Rancho Palos Verdes, California, USA. [1996 : 138] Let p∞ = 10
∞
1+
k=2
1 . Then clearly 1 < p < p∞ = Q h2
∞
≈
k=2
1 1+ 2 h
Now, ln R =
1.6714 and R =
∞
k=11
k=11
1 ln 1 + 2 < k
∞
10
1+
× R where Q =
1 . k2
1 ln 1 + 2 x
dx <
∞ dx
10
x2
=
1 . (Ed: This 10
36
is because f (x) = ln 1 + for t > 0). Therefore, R < e0.1
1 is strictly decreasing on (0, x2
∞) and ln(1 + t) < t
≈ 1.1052, and thus, p ∞ < (1.68)(1.11) = 1.8648 < 2.
1752 . [1992 : 175]
If A and B are positive integers and p is a prime such that p A, p2 p2 B, then the arithmetic progression
|
|
| A and
A, A + B, A + 2B, A + 3B, . . . contains no terms which are perfect powers (squares, cubes, etc.). Are there any infinite non-constant arithmetic progressions of positive integers, with no term a perfect power, which are not of this form? I. Solution by Margherita Barile, student, Universit¨ at Essen, Germany. [1993 L 149] The answer is in the affirmative. Let p be a prime, p > 2. Since 2 1 ( p 1)2 1 (mod p), there is an r satisfying 0 < r < p, such that a2 r p 2 3 (mod p) for all a = 0, . . . , p 1. Let A = p r , B = p . Then, for all k, we have
≡ −
≡
≡
−
A + kB = p2 r p + kp3 = p2 (r p + kp). The greatest power of p dividing A + kB is p 2 . Hence, if A + kB is a perfect power, it is a square. Then there is an integer s such that r p + kp = s2 . But then, by Fermat, s2
≡
r p
≡
r (mod p),
which provides a contradiction. Thus, the arithmetic progression A, A+B, A+2B, . . ., contains no perfect powers. But it is not of the given form. In fact, our proof shows how to construct infinitely many such arithmetic progressions. We give one example. For p = 3, one gets r = 2; then A = 72, and B = 27. II. Solution by Leroy F. Meyers, Ohio State University, Ohio, USA. [1993 : 150] For each non-negative integer m let x m = A + mB. The proof of the first statement of the problem is trivial. (In fact, it may not have been intended to be part of the problem.) Suppose that p is a prime and that p A, p2 A, and p2 B. Then xm A 0 (mod p 2 ), so that p x m and p2 xm . But if xm is a perfect k th power, then pk must divide xm , which is impossible if k > 1.
|
|
|
|
≡ ≡
|
Two cases of the converse are considered. If no prime dividing A occurs to a higher power in B than in A, then A is relatively prime to C = B/ gcd(A, B). By Euler’s theorem we have Aφ(C )
≡
1
(mod C ),
37
so that A1+φ(C )
≡
A (mod AC ).
but AC = AB/ gcd(A, B) = lcm(A, B), and thus, B AC . Hence,
|
A1+φ(C )
≡
A (mod B ),
and A 1+φ(C ) is the required perfect power in the arithmetic sequence. However, if A is divisible by a prime which occurs to a higher power in B, then there may be no perfect power in the sequence. For example, let A = 12 = 2 2 3 and B = 16 = 24 . If
·
12 + 16m = sk
for some positive integer k,
then s must be even, in which case 12 + 16m must be divisible by 2 k , which is impossible if k > 2. However, if k = 2 and s = 2t, then 3 + 4m = t2 , which also is impossible, since a perfect square must be congruent to either 0 or 1 modulo 4. Thus, there can be a non-constant perfect-power-free arithmetic sequence not of the specified form.
1434.
[1989 : 110] Proposed by Harvey Abbott and Murray S. Klamkin, University of Alberta, Edmonton, Alberta. It is known that (3m)!(3n)! , m!n!(m + n)!(n + m)!
(4m)!(4n)! , m!n!(2m + n)!(2n + m)!
(5m)!(5n)! m!n!(3m + n)!(3n + m)! are all integers for positive integers m, n. 1. Find positive integers m, n such that I (m, n) =
(6m)!(6n)! m!n!(4m + n)!(4n + m)!
is not an integer. 2. Let A be the set of pairs (m, n), with n m, for which I (m, n) is not an integer, and let A(x) be the number of pairs in A satisfying 1 n m x. Show that A has positive lower density in the sense that
≤
lim inf
x
→∞
A(x) > 0. x2
≤ ≤ ≤
38
Solution by Marcin E. Kuczma, Warszawa, Poland . [1990 : 184] The clue to (ii) is the observation that the set Q =
{(x, y) : 0 ≤ y ≤ x ≤ 1, 6x+6y
<
(1)
4x + y + 4y + x}
is non-empty ( here x is the greatest integer
Q =
{(x, y) : 4x + y ≥
2, 4y + x
≤ x); namely ≥ 1, x < 1/2, y
> 1/6
}
(2)
is the quadrangle with vertices 1 1 , 2 6
11 1 , 24 6
7 2 , 15 15
1 1 , 2 8
,
,
,
.
(There is no need to verify that the sets (1) and (2) are equal; it suffices for the sequel to take (2) as the definition of Q.) What follows is routine. Choose a prime p > 2 and suppose (m, n) is a lattice point in pQ =
{( px, py) : (x, y) ∈ Q}.
By (2)
6m 6n 4m + n 4n + m 2, < 3, < 1, p p p p Thus, p enters the denominator of I (m, n) with the exponent n < m,
≥
m n 4m + n 4n + m + + + p p p p
and numerator with the exponent
6m 6n + p p
≤
≥
≥
1.
(3)
3
2
(there are no higher order terms, since p 2 exceeds 6m). Consequently
(Z2 pQ)
p prime
∩
⊆
A.
The claim thus follows immediately: denoting by p(x) the greatest prime less than or equal to x, we have asymptotically p(x) x and thus, A(x) x2 as x
≥
A( p(x)) x2
≈
| ≈ Area Q ≥ |Z ∩ p(x)Q 2 x
→ ∞.
1 = 720
An example of (i) would be m = 11, n = 3, which satisfy (3) for p = 23. Editor’s comment . [1990 : 185] The proposers ask whether lim (A(x)/x2 ) exists. x
→∞
The problem of showing that
(5m)!(5n)! m!n!(3m + n)(3n + m)! is integral occurred in the 1975 U.S.A. Mathematical Olympiad (and was suggested by Klamkin).
39
3
Functions and Polynomials
299. [1976 : 298] If F 1
= ( r 2 + s2
F 2
+ 4rt(x2 + y 2 ), = 2rs(x2 y 2 2xy) + (r2
F 3
+ 4st(x2 + y 2 ), = 2rt(x2 y 2 2xy)
−
− 2t2)(x2 − y2 − 2xy) − 2rs(x2 − y2 + 2xy)
−
− −
−
− −
− s2 − 2t2)(x2 − y2 + 2xy)
− 2st(x2 − y2 + 2xy)
+ (r2 + s2 + 2t2 )(x2 + y 2 ),
show that F 1 , F 2 and F 3 are functionally dependent and find their functional relationship. Also, reduce the five-parameter representation of F 1 , F 2 , and F 3 to one of two parameters. Solution de F.G.B. Maskell, Coll`ege Algonquin, Ottawa Ontario. [1978: 170] Posons A = x(r B = x(s
− t) + y(s + t), − t) − y(r + t).
On v´erifie, avec un peu de patience, que
F 3 + F 2 = 2A(A F 3
− F 1
− B),
F 3
= 2A(A + B),
− F 2
= 2B(A + B),
F 3 + F 1 =
−2B(A − B).
(1)
On a donc (F 3 + F 2 ) (F 3
− F 2) + (F 3 + F 1) (F 3 − F 1)
= 0,
ce qui am`ene F 12 + F 22 = 2F 32 .
(2)
C’est la relation fonctionnelle recherch´ee. On verra que les deux param`etres A et B suffisent pour repr´esenter les trois fonctions donn´ees. En effet, de 2F 3
− F 1 + F 2
= 4A2 et 2F 3 + F 1
on obtient
−F 1 + F 2 et
= 2(A2
− F 2
= 4B 2 ,
− B2)
(3)
F 3 = A2 + B 2 .
(4)
Portons maintenant dans (2) les valeurs de F 2 et F 3 de (3) et (4); il r´esulte F 12 + 2 A2
− B2
F 1 + A4
− 6A2B2 + B4
= 0,
40
d’o` u F 1 =
−A2 ± 2AB + B2. Ici il faut prendre F 1 = −A2 − 2AB + B 2 ,
car l’autre choix, avec la valeur de F 2 qui d´ecoule alors de (3), ne v´erifie aucune des relations (1). La repr´esentation recherch´ee est donc F 1 =
−A2 − 2AB + B2 ,
F 2 = A2
− 2AB − B 2,
F 3 = A2 + B 2 .
Editor’s comment.[1978 : 171] The proposer mentioned that the problem arose in applying the method of Desboves in obtaining the general solution of the Diophantine equation F 12 + F 22 = 2F 32 from the knowledge of one particular solution. An analogous but more complicated set of equations would arise if we started with the general homogeneous quadratic Diophantine equation in n variables and one particular solution.
254. [1976 : 155] 1. If P (x) denotes a polynomial with integer coefficients such that P (1000) = 1000,
P (2000) = 2000,
P (3000) = 4000,
prove that the zeros of P (x) cannot be integers. 2. Prove that there is no such polynomial if P (1000) = 1000,
P (2000) = 2000,
P (3000) = 1000.
Solution by Leroy F. Meyers, Ohio State University, Columbus, Ohio, USA. [1978 : 50] A generalization yielding both parts is proved. Let a, b, c, d, k be integers, with k = 0. Suppose P is a polynomial with integral coefficients such that
P (d
− k)
= ak,
P ( d) = bk,
and
P (d + k) = ck.
If we define Q by Q(x) = P (x+d), then Q is a polynomial with integral coefficients and Q( k) = ak, Q(0) = bk, and Q(k) = ck.
−
Now the polynomial Q 1 of lowest degree such that Q1 ( k) = ak,
−
Q1 (0) = bk,
and
Q1 (k) = ck
is given, by the Lagrange interpolation formula or otherwise, by Q1 (x) =
(a
− 2b + c)x2 + (c − a)kx + 2bk2 . 2k
41
Hence, the polynomial Q
− Q1 has zeros at −k, 0, k, and thus, Q(x) − Q1 (x) = (x + k)x(x − k)R(x)
for some polynomial R, or 2kQ(x)
− (a − 2b + c)x2 − (c − a)kx − 2bk2 = x3
− k2 x
(1)
S (x),
where S = 2kR. Now S , being the quotient of a polynomial with integral coefficients by a polynomial with integral coefficients and leading coefficient 1, must itself have integral coefficients. Equating the coefficients of x2 on the two sides of (1) yields 2kq
− (a − 2b + c)
−k2s,
=
where q and s are the coefficients of x 2 in Q(x) and x in S (x), respectively, and it follows that a
− 2b + c
is divisible by k.
(2)
If c a, and hence, c + a, is even, then slightly more can be proved. Instead of (1) we now have
−
kQ(x)
− a − 2b2 + c x2 − c −2 a kx − bk2
=
x3
− k2x
S (x),
(1’)
where now S = kR. Then, as before, S has integral coefficients and, after comparing the coefficients of x 2 , we obtain kq
− a − 2b2 + c
=
−k2s,
and thus, a
− 2b + c 2
is divisible by k.
(2’)
Note that (2) and (2’) are independent of d, which can therefore be any integer. In part (a) we have a = 1, b = 2, c = 4, so that a k is a divisor of 1, that is, k = 1.
±
− 2b + c = 1. Hence, by (2),
In part (b) we have a = 1, b = 2, c = 1, so that a−22b+c = 1. Since c a is even, we can now use (2’) to conclude that k is a divisor of 1, that is, k = 1.
−
−
− ±
Since k = 1000 in both parts of the proposal, each part yields a contradiction. Hence, there is no polynomial with integral coefficients which satisfies either of the given conditions, and thus, none of the zeros of P can be integers; also, all of them must be integers. (For an analogous situation, see Problem 138 [1976 : 157]).
42
1423 . [1989 : 73] Given positive integers k, m, n, find a polynomial p(x) with real coefficients such that (x 1)n ( p(x))m xk .
−
|
−
What is the least possible degree of p (in terms of k, m, n)? Solution by Robert P. Israel, University of British Columbia, Vancouver, BC . [1990 : 145] If p is a polynomial in x, pm xk is divisible by (x 1)n if and only if
−
pm = xk + O((x
− 1)n)
as
− x → 1.
[Editor’s note : This means that p m = x k + r(x), where
|r(x)| ≤ M |x − 1|n
(1)
for some constant M and for x sufficiently close to 1.] This is true if and only if p = xk/m + O((x
− 1)n).
[Editor’s note : Len Bos contributes the following elaboration for the editor, who regrets not paying more attention as a student during analysis class. We have p = xk/m +
1+
r(x) xk
1/m
−1
xk/m .
Applying the Mean Value Theorem to the function f (t) = (1 + t)1/m , t > yields that 1 (1 + t)1/m 1 = t (1 + c)1/m −1 m k for some c, c < t , so that with t = r(x)/x we get
|
− |
(2)
−1,
| |·
| | | |
r(x) xk
1+
1/m
−
1 =
|r(x)| · xk
1 (1 + c)1/m −1 m
for some c between 0 and r(x)/xk . For x sufficiently close to 1 this means
r(x) 1+ k x
1/m
−
1 xk/m
≤ K |x − 1|n
for some constant K , by (1). Thus, from (2) we get the result.] This means that p consists of the terms of the Taylor series of xk/m about x = 1 up to order (x 1)n−1 , plus any combination of higher powers of x 1. That Taylor series is
−
−
n 1
1+
− k/m(k/m − 1) · ·· (k/m − j + 1)
j =1
j!
(x
− 1)j .
(3)
43
If k is not divisible by m, all coefficients of the Taylor series are non-zero, so that the least degree of p is n 1. If k is divisible by m, the coefficients for j > k/m are zero, so that the least degree of p is min(n 1,k/m).
−
−
Editor’s comment . [1990 : 146] It appears that the series (3) can also be written as n i ( 1)n−1−i − k(k − m)(k − 2m) · ·· (k − (n − 1)m) i xi ; n − 1 (n − 1)!m k im − i=0
−
−
n 1
can any reader supply a proof? II. Comment by Rex Westbrook, University of Calgary, Calgary, Alberta. [1991 : 83] This is in response to the editor’s request for a proof that the published answer n 1 k k m(m
−
1+
− 1) · ·· ( km − j + 1) (x − 1)j j!
j =1
can be written as k k m(m
− 1) · ·· ( km − n + 1) n−1 (−1)n−1−i n − 1 k (n − 1)! i − i m i=0
xi .
Set the former expression equal to f n (x) and the latter equal to gn (x). The asked equality obviously holds if k/m = t n 1, where t is an integer, because then both expressions reduce to x t . Otherwise, consider
≤ −
d dx
k
n−1
+
j !
j=1 k
= x−( m +1) k +1) −( m
= x
− − − k
m k
m
k
= x−( m +1)
−
j=1
k +1) −( m
+
+
n−1 k k ( m m j=1 n−1 k m j=1 n−1 k m
j−1 −k/m
j (x − 1)
x
− 1) · · · ( km − j + 1) j !
· · · ( km − j + 1) j !
−
k m
k
−
−
+
j !
k k ( m m
m
k j −m −1
(x − 1) x
(x − 1)j−1 jx −
− j
· · · ( km − j + 1) (x − 1)j−1 m j − 1)! ( j=1 k k · · · ( m − j + 1)( mk − j ) m (x − 1)j k
n−1
= x
k m
− x−( +1) m k k k − · · · − j + 1) ( 1) ( m m m
[x−k/m f n (x)] =
− 1) · · · ( km − (n − 1)) (x − 1)n−1 (n − 1)!
k m
(x − 1)
(x − 1)j + j (x − 1)j−1
44
and
− 1) · · · ( km − (n − 1)) d n−1 (−1)n−1−i n−1 i gn (x)] = [x k dx dx i=0 (n − 1)! i − m n−1 k k k n − · · · − − ( 1) ( ( 1)) n−1 m = m m (−1)x−( +1) (−1)n−1−i xi i (n − 1)! i=0 k k k n − − · · · − − ( 1) ( ( 1)) m m m = x−( +1) (x − 1)n−1 . (n − 1)! d
k k ( m m
−k/m
k m
k m
i−k/m
x
Therefore, d x−k/m (gn (x) dx
− f n(x))
= 0,
so that x−k/m (gn (x)
− f n(x)) =
constant.
If either k/m is not an integer, or is an integer n, then since gn f n is a polynomial (of degree at most n 1) the above constant must be 0; that is, gn f n .
≥
−
≡
−
2014. [1995 : 52] (a) Show that the polynomial 2 x7 + y 7 + z 7
has x + y + z as a factor.
7xyz x4 + y 4 + z 4
−
(b) Is the remaining factor irreducible (over the complex numbers)? I. Solution to (a) by Jayabrata Das, Calcutta, India . [1996 : 45] Let f (x,y,z) = 2 x7 + y 7 + z 7 7xyz x4 + y 4 + z 4 . If we can show that f (x,y,z) = z when x + y + z = 0, we are done.
−
We know, for x + y + z = 0, that x 3 + y 3 + z 3 = 3xyz. Thus, x7 + y 7 + z 7 + x3 y 4 + x3 z 4 + y 3 z 4 + y 3 x4 + z 3 y 4 + z 3 x4 = x3 + y 3 + z 3 x4 + y 4 + z 4
= 3xyz x4 + y 4 + z 4 so that
x7 + y 7 + z 7
=
3xyz x4 + y 4 + z 4
−x3y4 − x3 z4 − y3z4 − y3x4 − z3y4 − z3x4
45
Therefore, f (x,y,z) = 2 x7 + y 7 + z 7
7xyz x4 + y 4 + z 4
− − − − − − − − − − − −
= 6xyz x4 + y 4 + z 4
2 x3 y 4 + x3 z 4 + y 3 z 4 + y 3 x4 + z 3 y 4 + z 3 x4 7xyz x4 + y 4 + z 4
xyz x4 + y 4 + z 4
=
2x3 y 3 (x + y)
2y 3z 3 (y + z)
− 2z3x3 (z + x)
xyz x4 + y 4 + z 4 + 2x3 y 3 z + 2xy 3 z 2 + 2x3 yz 3
=
xyz x4 + y 4 + z 4
= = Since x 2 + y 2 + z 2 =
2
xyz
2
x +y +z
2x2 y 2
2 2
2y 2 z 2
2 2
− 2z 2x2 2 2
2 2
4 x y +y z +z x
−2(xy + yz + zx), we now have that
f (x,y,z) =
2
−
2 2
−xyz 4 (xy + yz + zx) 4 x y −4xyz 2x2yz + 2xy2z + 2xyz 2 −8xyz xyz(x + y + z) = 0.
= =
2 2
2 2
+y z +z x
.
II. Solution to (a) by Cyrus Hsia, student, University of Toronto, Toronto, Ontario. [1996 : 45] Consider the sequence an = xn + y n + z n . The characteristic equation with roots x, y , z, is a3 Aa2 + Ba C = 0,
−
−
where A = x + y + z, B = xy + yz + zx and C = xyz. The sequence an follows the recurrence relation:
{ }
an+3 = A an+2
− B an+1 + C an.
Now, we have a0
= x0 + y 0 + z 0 = 3,
a1 a2
= x1 + y 1 + z 1 = A, = x2 + y 2 + z 2 = (x + y + z)2
− 2(xy + yz + zx)
= A2
From the recurrence relation, we see: a3
= A a2 B a1 + C a 0 = A3 2AB AB + 3C =
− −
−
A k3 + 3C, where k 3 is some term in x, y and z
− 2B.
46
Similarly a4 a5
= A k4 + 2B 2 , where k 4 is some term in x, y and z , = A k5 5BC, where k 5 is some term in x, y and z,
a6 a7
= = A k7 + 7B 2 C, where k 7 is some term in x, y and z .
− A k6 − 2B 3 + 3C 2 , where k 6 is some term in x, y and z ,
Thus, 2 x7 + y 7 + z 7
= = =
7xyz x4 + y 4 + z 4
− − −
2a7 7C a 4 a A k7 + 7B 2 C A k,
7C A k4 + 2B 2
where k is some term in x, y and z ; that is, x + y + z divides 2 x7 + y 7 + z 7 7xyz x4 + y 4 + z 4 .
1110 . [1986 : 13]
−
How many different polynomials P (x1 , x2 , . . . , xm ) of degree n are there for which the coefficients of all the terms are 0’s or 1’s and P (x1 , x2 , . . . , xm ) = 1 whenever x1 + x2 +
· ·· + xm
= 1?
Editor’s comment . [1987 : 170] There has been only one response to this problem, and it was incorrect. It did note the easy case m = 1, where P (x) = xn is the only solution for each n. Equally easy is the case n = 1 which has unique solution P (x1 , x2 , . . . , xm ) = x1 + x2 + + xm for each m. I would not like to see this problem abandoned at this point. Can anyone find all such p olynomials for any other values of m and/or n? Can anyone show that there is at least one such polynomial for each m and n?
· ··
I. Partial solution by Len Bos and Bill Sands, University of Calgary, Calgary, Alberta . [1988 : 13] Let f (n, m) be the required number of polynomials. We will investigate the case m = 2 and will show that f (n, 2)
≥
1 2n , n+1 n
the n–Catalan number. Let P (x, y) be a polynomial of degree n with the required properties. Then P (x, y) = 1 whenever x + y = 1, so that it must be true that P (x, y) = (x + y
− 1) q (x, y) + 1
(1)
for some polynomial q (x, y) with integer coefficients. We shall count all those possible q (x, y) whose coefficients are also all 0 or 1.
47
Let q be such a polynomial. Then its terms are monomials of the form xi y j , where i, j 0, 1, . . ., n 1 and i + j n 1. We will identify the collection of these monomials with the corresponding subset of lattice points
∈ {
− }
Rq =
≤ −
{(i, j)|xiyj is a monomial in q }.
Thus, Rq is a subset of (i, j) Z 2 0 i, 0 j, i + j n 1 and contains at least one (i, j) with i + j = n 1. The next two lemmas establish important properties of R q .
{
∈ | ≤ −
≤
≤ − }
Lemma 1. If (i, j) Rq where i + j > 0, then either (i 1, j) Rq or (i, j 1) Rq (or both). In particular, if (i, 0) Rq then (i 1, 0) Rq for i > 0, and similarly for (0, j).
∈
− ∈
∈
−
−
∈
∈
Proof . If (i, j) Rq and i + j > 0 then xi y j is a monomial in q . Thus, (x + y 1)q (x, y), when multiplied out, will contain a term xi y j .
∈
−
−
By (1), it must therefore also contain at least one term + xi y j , which can only happen if xi−1 y j or xi y j−1 were monomials in q ; that is, if (i 1, j) or (i, j 1) Rq .
−
− ∈
In terms of lattice points, this lemma says that if a lattice point is in R q , then at least one of its neighbours to the left of or below it must also be in Rq . Our other lemma is a sort of converse.
Lemma 2 . If (i
(i − 1, j )
(i, j )
(i, j − 1)
− 1, j) ∈ Rq and (i, j − 1) ∈ Rq , then (i, j) ∈ Rq .
Proof . We have that xi−1 y j and xi y j−1 are both monomials in q . Then (x + y 1)q (x, y) when multiplied out will contain two terms x i y j . By (1), it must also contain a term xi y j , which implies that (i, j) Rq .
−
−
∈
Now suppose (i, j) Rq . By applying Lemma 1 repeatedly, we obtain a descending path of lattice points in R q from (i, j) to (0, 0). By always moving left from a lattice point rather than down, whenever we have a choice, we obtain what we call the left path of (i, j). Similarly by moving down instead of left whenever possible, we obtain the right path of (i, j). The diagram shows a possible left path of (4, 3). All lattice points on the path are in R q , but the position of the path tells
∈
48
us that (2, (2, 3) and (2, (2, 2) are not are not in in R q . . .. ............................................................................. .... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ............................................................................... ............................................................................ ........................................................................... .. ................. . . . . . . . . . . . . . . . . . . . . . . . . .
(2, 3)
(4, 3)
(2, 2)
(0, 0)
Clearly the left path and right path of (i, (i, j ) do not cross, although they may meet (and do, at their end-points at least). (i, j ) . . . . . . . . . . . . . . . ... . .. .................................. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ... . .. .................................... .................................. .... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .................................... ................................. ... .... . ... . . . . . . . . . . . . . . . . . . . . . . . . . . . . ... . .. .................................... .................................... .................................. ....... . . . . . . . . . . . . . . . . . . . . . . . . . . ... ... ..................................... .................................... .................................. .......
(k, j )
(k, )
(0, 0)
Let (i, (i, j ) and (k, (k, ) be in R in R q , where we assume that i < k and j and j < . . Consider the left path of (i, ( i, j ) and the right path of (k, ( k, ). Extend them to paths beginning at (k, ( k, j ) by adding horizontal and vertical edges, respectively. The extended paths then enclose a region of lattice points. Clai Claim: m:
Ever Everyy latti latticce point oint insi inside de this this region gion is in R in Rq .
This follows by repeated applications of Lemma 2, starting at the bottom left of the region and working up and to the right. Now it can easily be seen that Rq must coincide with the region of lattice points bounded by the left and right paths of some lattice point (i, ( i, j ), where i where i + j = j = n n 1. Furthermore we claim than any such region corresponds to a polynomial q (x, y ) such that P that P ((x, y ), defined by (1), is a polynomial satisfying the problem. We need only show that P ( P (x, y ) has coefficients 0 or 1. Multiplying out (x ( x + y 1)q 1)q (x, y ), we need only show that any negative term xi y j , i + j + j > 0, is offset by at least one term +x +xi y j , and that if two terms xi y j occur then also a term xi y j will occur. But this follows from the construction of the region much as in the proofs of Lemmas 1 and 2.
−
− −
−
(i, j ) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ... ... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ................................. .. ................................. .............................. ........
(i, j )
P
−→
P r
(0, 0)
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ... ... . . . . . . . . . . . . . . . . . . . . . . . . . .................................. ... .
(0, 0)
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ..................................................................................................... .......
P
P r
49
Thus, to count all the polynomials q polynomials q (x, y ) we must count the number of pairs of lattice paths P and P r from (0, (0, 0) to (i, (i, j ), i + j + j = n 1, which do not cross and have length n 1. By movin moving g path path P r one unit to the right and one unit down, adding in new common end-points as shown, and moving both paths to start at (0, (0 , 0) again, we see that such pairs of paths correspond to those pairs of paths from (0, (0, 0) to (i, (i, j ), where i where i + j = j = n n + 1, whic which h do not meet (except (except at their end-points), and which have length n length n +1. The number of such pairs of paths, over all choices of i, i , j satisfying i satisfying i 0, j 0, j 0, i 0, i + j = j = n + 1, is known to be the Catalan Catalan number 1 2n n+1 n
−
−
≥
≥
(see J. Levine, Note on the number of pairs of non-intersecting routes, Scripta Mathematica 24 Mathematica 24 (1959) 335–338). This number is then a lower bound for f ( f (n, 2). Unfortunately it is not the exact answer, since putting q (x, y ) = 1 + x + y + 2xy 2 xy + + x2 y + xy 2 + x2 y 2 (which has a coefficient not equal to 0 or 1) into (1) yields P ( P (x, y ) = x2 + y 2 + x2 y + xy 2 + x3 y + xy 3 + x2 y 2 + x3 y 2 + x2 y 3 , a polynomial with all coefficients 0 or 1. We do believe, however, that f ( f (n, 2) can 2) can be calculated, and, as a possible first step, make the following conjecture: any q (x, y ) suitable for (1) has all coefficients 0, 1 or 2. II. II. Partial solution by P. Penning, Delft, the Netherlands . [1988 : 16] (Adapt (Adapted ed by the editor to refer to I above.) We show that f ( f (n, m) mn−1 ,
≥
thus, answering the editor’s request [1987 : 170] for a proof that f ( f (n, m) each n each n and m. m .
≥ 1 for
A special case of the allowable “regions” in part I part I is that of a single path from (0, (0, 0) to (i, (i, j ), where i + j + j = n 1. A similar argumen argumentt to that in I shows more generally that if P P is a path of length n 1 from (0, (0, 0, . . . , 0) to (i (i1 , i2 , . . . , im) in Z m , where m i = n 1, then the lattice points on P P will correspond to j =1 j monomials whose sum is a polynomial q polynomial q (x1 , . . . , xm ) such that
−
−
−
P ( P (x1 , . . . , xm ) = (x1 + . . . + xm
− 1) q (x1, . . . , xm ) + 1
has all coefficients 0 or 1. To construct such a path, we merely choose a sequence of n n 1 elements from x from x 1 , . . ., x m , repetition allowed, each corresponding to one of the m “directions” the path can take (starting at (0, (0 , . . . , 0)). 0)). The numbe numberr of n−1 these sequences is m .
−
Example : m = 5, n 5, n = = 4. Choose sequence x sequence x 2 , x 4 , x3 . Then q (x1 , x2 , x3 , x4 , x5 ) = 1 + x2 + x2 x4 + x2 x4 x3 ,
50
so that )(x1 + x2 P ( P (x1 , . . . , x5 ) = (1 + x2 + x2 x4 + x2 x4 x3 )(x +x3 + x4 + x5 1) + 1 = x1 + x3 + x4 + x5 + x2 (x1 + x2 + x3 + x5 )
−
+x2 x4 (x1 + x2 + x4 + x5 ) +x2 x3 x4 (x1 + x2 + x3 + x4 + x5 ).
1283. [1987 : 289] Show that the polynomial P ( P (x,y,z) x,y,z ) = (x2 + y 2 + z 2 )3
− (x3 + y3 + z3)2 − (x2y + y2z + z 2x)2 − (xy2 + yz 2 + zx2)2
is non-negative for all real x, x , y , z . Solution by Jorg Herterich, student, Winnenden, Federal Republic of Germany . Germany . [1988 : 306] This can be seen by writing the expression another way: (x2 + y 2 + z 2 )3 (x3 + y 3 + z 3)2 (x2 y + y 2 z + z 2 x)2 (xy 2 + yz 2 + zx 2 )2 =
= =
−
−
−
2x4 y 2 + 2x 2x4 z 2 + 2y 2y 4 x2 + 2y 2y 4 z 2 + 2z 2z 4 x2 + 2z 2z 4 y 2 + 6x2 y 2 z 2 2x3 y 3 2y 3 z 3 2x3 z 3 2x3 z 2 y 2x3 y 2 z 2y 3 x2 z 2y 3 z 2x 2z 3 x2 y 2z 3 y 2 x (x2 y 2 + y 2 z 2 + x2 z 2 )(2x )(2x2 + 2y 2y 2 + 2z 2z 2 2xy 2yz 2xz) xz )
−
− −
−
−
−
−
−
− − − − (x2 y 2 + y 2 z 2 + x2 z 2 )((x )((x − y )2 + (y (y − z )2 + (x (x − z )2 ).
It is obvious that this is non-negative for all real x real x,, y , z .
51
4
Expressions and Identities
1304. [1988 : 12] If p, q , r are the real roots of x3
− 6x2 + 3x + 1
= 0,
determine the possible values of p2 q + q 2 r + r2 p, and write them in a simple form. Solution by Sam Baethge, Science Academy, Austin, Texas, USA. [1989 : 30] Let A = p2 q + q 2 r + r2 p, B = p2 r + q 2 p + r2 q, the only two possible values of expressions of the given type. We also have p + q + r = 6, pq + qr + rp = 3, pqr =
−1.
In the equations that follow, all summations are symmetric over p, q and r.
p2 q + 3 pqr =
1 8 = ( p + q + r)( pq + qr + rp) = or
216 = ( p + q + r)3 =
p3 = 216
p3 q 3 = 27
AB
=
p2 q + 6 pqr
p3 q 3 + 3
p2 q = 21
p3 q 2 r + 6 p2 q 2 r2
− 3(−1)(21) =
p3 q 3 + 3 p2 q 2 r2 = pqr
= ( 1)(159) + 87 =
−
p3 + 3
− 6(1) − 3 pqr
p4 qr +
−72.
Using (1) and (2), A and B are the roots of y2
(1)
− 3(21) − 6(−1) = 159.
2 7 = ( pq + qr + rp)3 = or
−3
p2 q = 21.
A+B =
or
p2 q
− 21y − 7 2 = 0, so that the possible values are 24 and −3.
p3 + 84 + 3
84.
(2)
52
287. [1976 : 251] Determine a real value of x satisfying
2ab + 2ax + 2bx
− a2 − b2 − x2
=
if x > a, b > 0.
− − ax
a2 +
bx
b2
Composite of the solutions received from Gali Salvatore, Perkins, Qu´ ebec; and Murray S. Klamkin, University of Alberta, Edmonton, Alberta. [1978 : 135] The restriction 0 < a, b < x is equivalent to
√ a, √ b
0 <
√ x.
<
(1)
The radicand on the left of the given equation can be factored by inspection, giving
√ √ √ √ √ √ √ √ √ √ √ √ − − − + + + + + a
b
x
b
x
=
a
x
a
b
a
b
x
√ ax − a2 + √ bx − b2.
(2)
Now (1) ensures that the right side of (2) and the first three factors of the radicand on the left are all positive; hence, for a solution to exist the fourth factor must also be positive, and we require
√ x
<
√ a + √ b.
(3)
Thus, with (1) and (3) the given equation is equivalent to the one obtained by squaring both sides. This procedure gives, after simplification, x(a + b
− x)
√
= 2 ab =
− (x
a)(x
− b) −
√ − a − b) √ . (x − a)(x − b) + ab 2x ab(x
√
ab
(4)
One solution of (4) is clearly x = a + b; and it is the only one since x = a + b makes one side of (4) positive and the other side negative.
Since x = a + b also satisfies both (1) and (3), it is the unique solution to the given equation. Editor’s comment.[1978 : 136]
√
√
A .................................................................................x M x C ........................................................................................................................................................................... ........... . ... .. ..... . ... .. ...... ... .. ... ...... ..... .. ... . . ...... ... . . . . . . . ... . ...... ... .. ..... .... .. ..... . ... ..... ... .. . ...... .... ..... .. . . ... . . . . .... ... .. ... ..... ... ..... .. ... ...... .. . .... ...... .. ...... .... . . . . . . ... ...... .. .... . ..... .... . .. ...... ... ........... .... . .......
√
2 a
√
x
B
√
2 b
The restrictions (1) and (3) ensure that a, b, x (or any constant multiple thereof) are the lengths of the sides of a triangle. In fact, the proposer pointed out that the given equation can be interpreted geometrically by the following area relationship (see figure):
√ √ √
53
|ABC |
= ABM + M BC .
|
| |
|
This is most easily verified from (2) by Heron’s formula. Thus, the given equation and its solution could be used to give us another proof, if one were needed, of the following theorem: If the mid-point of longest side of a triangle is equidistant from the three vertices, then the triangle is right–angled. Conversely, this theorem itself can be used to provide an unexpected solution to the given equation.
1594. [1990 : 298] Express x41 + x42 + x43 + x44
− 4x1x2x3 x4
as a sum of squares of rational functions with real coefficients. (By the AM–GM Inequality, this polynomial is non-negative for all real values of its variables, and therefore, by a theorem of Hilbert, it can be so expressed.) Solution by Iliya Bluskov, Technical University, Gabrovo, Bulgaria. [1992 : 22] x41 + x42 + x43 + x44 4x1 x2 x3 x4 = x41 + x42 2x21 x22 + x43 + x44
−
= (x21
−
− x22)2 + (x23 − x24)2 +
− 2x23x24 + 2x21x22 + 2x23x24 − 4x1 x2x3 x4 2 √ 2(x1 x2 − x3 x4 ) .
Editor’s note . It is known more generally that, for any even n, an1 + an2 +
· · · + ann − na1a2 · ·· an
can be expressed as a sum of squares; see p. 55, section 2.23 of Hardy, Littlewood, and P´ olya, Inequalities , Cambridge Univ. Press.
1522. [1990 : 74] Show that if a, b, c, d, x, y > 0 and xy = ac + bd, then
x ad + bc = , y ab + cd
abx cdx ady bcy + = + . a + b + x c + d + x a+d+y b+c+y
I. Solution by Seung-Jin Bang, Seoul, Republic of Korea. [1991 : 127] Note that, from the given expressions for xy and x/y, ab(c + d + x) + cd(a + b + x) = ad(b + c + y) + bc(a + d + y)
54
and x(a + d + y)(b + c + y) = ((a + d)x + ac + bd)(b + c + y) = (ac + bd)y + (a + d)(b + c)x +(b + c)(ac + bd) + (a + d)(ac + bd) = (ac + bd)x + (a + b)(c + d)y +(c + d)(ac + bd) + (a + b)(ac + bd) = ((a + b)y + ac + bd)(c + d + x) = y(a + b + x)(c + d + x). It follows that x
ab cd + a + b + x c + d + x
= y
ad bc + a + d + y b + c + y
.
II. Combination of partial solutions by Wilson da Costa Areias, Rio de Janeiro, Brazil, and Francisco Bellot Rosado, I.B. Emilio Ferrari, Valladolid, Spain. [1991 : 127] It is well known (see p. 111, no. 207 of N. Altshiller-Court, College Geometry ) that it is possible to construct a cyclic quadrilateral ABCD of sides AB = a, BC = b, CD = c, DA = d [provided that a < b + c + d, etc. — Ed.] and also that, from Ptolemy’s Theorem, its diagonals x = AC , y = BD satisfy xy = ac + bd,
x ad + bc = . y ab + cd
Let R be the circumradius of ABCD, and let F 1 , F 2 , F 3 , F 4 and s1 , s2 , s3 , s4 be the areas and semiperimeters of the triangles ABC , BC D, CDA, DAB, respectively. Then abx cdx bcy ady + = + a+b+x c+d+x b + c + y a + d + y is equivalent to F 1 4R F 3 4R F 2 4R F 4 4R + = + , 2s1 2s3 2s2 2s4
·
·
·
·
or r1 + r3 =
F 1 F 3 F 2 F 4 + = + = r2 + r4 , s1 s3 s2 s4
where r1 , r2 , r3 , r 4 are, respectively, the inradii of the above triangles. The relation r1 + r3 = r2 + r4 is true and has been shown by H. Forder in “An ancient Chinese theorem”, Math Note 2128, p. 68 of Mathematical Gazette 34, no. 307 (1950). It was also part (b) of Crux 1226 [1988 : 147].
55
830. [1983 : 80] Determine all real λ such that
|z1z1(z3 − z2) + z2z2(z1 − z3) + z3z3(z2 − z1) − iz1λ| = |(z2 − z3 )(z3 − z1 )(z1 − z2 )| , where z 1 , z 2 , z3 are given distinct complex numbers and z 1 = 0.
Solution by Murray S. Klamkin, University of Alberta, Edmonton, Alberta (revised by the editor). [1984 : 202] Consider the triangle (assumed to be non-degenerate) whose affixes in the complex plane are z1 , z2 , z3 , oriented in the sense of increasing subindices, and let ∆, R, ω denote its signed area, circumradius, and affix of its circumcentre, respectively. It is known that ω =
z1 z 1 (z3
and R =
− z2) + z2z2 (z1 − z3) + z3z3(z2 − z1) −4i∆ |(z2 − z3)(z3 − z1)(z1 − z2)| . 4|∆|
The given equation is therefore, equivalent to
z 1 λ ω + = R. 4∆
(1)
It is clear that (1) is satisfied if and only if 1λ − z4∆
= z
(2)
for some point z on the circumcircle, and then that λ is real if and only if z = z 1 or z = z 4 where z 4 is the point where the line through the origin and z 1 meets the circumcircle again. For z = z 1 , we obtain from (2) λ = λ1 =
−4∆.
To obtain the value of λ when z = z 4 in (2), we first note that the power of the origin with respect to the circumcircle is given, in both magnitude and sign, by the real number z 1 z4 = ω 2 R2 . Hence, z = z 4 in (2) when
||−
λ = λ2 =
−
4∆ ω 2 R2 . z1 2
| | − | |
Editor’s comment [1984 : 203] Henderson found the values of λ 1 and λ 2 explicitly in terms of z 1 , z 2 , z 3 : λ1 = i
z 1 (z2
where the sums are cyclic.
− z3),
λ2 =
i
|z1|2
z 1 z 2 z3 (z2
− z1),
56
1996. [1994 : 285] (a) Find positive integers a 1 , a 2 , a 3 , a 4 such that (1 + a1 ω)(1 + a2 ω)(1 + a3 ω)(1 + a4 ω) is an integer, where ω is a complex cube root of unity. (b) Are there positive integers a 1 , a 2 , a 3 , a 4 , a 5 , a 6 so that (1 + a1 ω)(1 + a2 ω)(1 + a3 ω)(1 + a4 ω)(1 + a5 ω)(1 + a6 ω) is an integer, where ω is a complex fifth root of unity? I. Solution by Kee-Wai Lau, Hong Kong . [1995 : 311] (a) We have (1 + 2ω)4 = 9. [Editor’s note . Since ω = 1/2 1 + 2ω = 3 i.]
−
±√
± (√ 3/2)i, we have
(b) We shall answer in the negative. We first show that if A, B, C , D are integers such that A + Bω + Cω 2 + Dω 3 = 0, (1) then A = B = C = D = 0. We may assume that ω = e 2πi/5 = cos72◦ + i sin72◦, as the proofs for other choices of ω are similar. The real part and imaginary part of the left hand side of (1) both vanish and thus, A + B cos72◦
− (C + D)cos36◦
= 0
(2)
and B sin72◦ + (C
− D)sin36◦ = 0. (3) From (3) we obtain 2B cos36◦ + (C − D) = 0. Since cos36◦ is irrational [in √ fact, cos36◦ = ( 5 + 1)/4 — Ed .], B = 0 and C = D. Thus, (2) becomes A − 2C cos 36◦ = 0 and thus, A = C = 0. [Editor’s note . Alternatively, this result 2 3 4 is true because the polynomial 1 + z + z + z + z is irreducible and thus, must be the polynomial of smallest degree (with integer coefficients) that has ω as a root. See the (first) Editor’s Note in Solution III below.]
Now suppose, on the contrary, that the answer is in the affirmative. Using the fact that ω 4 = (1 + ω + ω 2 + ω 3 ) and ω 5 = 1, we see that the given product equals P + Qω + Rω 2 + T ω 3, where P , Q, R, T are integers and R = S 2 S 4 , T = S 3 S 4 where S k stands for the k th elementary symmetric function of a 1 , a 2 , . . ., a6 , k = 1, 2, 3, 4. Using the above result we see that R = T = 0 and thus, S 2 S 3 = 0. However, since ak 1 we have
−
−
−
−
≥
a1 a2 a1 a5 a2 a4 a3 a4 a4 a5
≤ ≤ ≤ ≤ ≤
a1 a2 a3 ,
a1 a3
a1 a5 a6 ,
a1 a6
a2 a4 a5 ,
a2 a5
a3 a4 a5 ,
a3 a5
a4 a5 a6 ,
a4 a6
≤ ≤ ≤ ≤ ≤
a1 a3 a4 ,
a1 a4
a1 a2 a6 ,
a2 a3
a2 a5 a6 ,
a2 a6
a2 a3 a5 ,
a3 a6
a2 a4 a6 ,
a5 a6
≤ ≤ ≤ ≤ ≤
a1 a4 a5 , a2 a3 a4 , a2 a3 a6 , a3 a4 a6 , a3 a5 a6 .
57
[Editor’s note . If the editor may put on his combinatorial hat for a minute, what is going on here is just a demonstration of the known fact that there is a “complete matching” from the 2–element subsets to the 3–element subsets of the set 1, 2, 3, 4, 5, 6 ; that is, a one-to-one function from the 2–element subsets to the 3–element subsets such that each 2–element subset is mapped to a 3–element subset containing it. In fact there is a complete matching from the k–element subsets of an n–element set to the (k + 1)–element subsets whenever k < n/2. See, for example, Corollary 13.3 on page 688 of [2], or (for the whole story and more) Chapters 1 to 3, especially Exercise 2.4 on page 23, of [1].] Therefore,
{
}
S 2
− S 3 ≤ −(a1a2a4 + a1a2a5 + a1a3 a5 + a1a3a6 + a1a4a6)
< 0,
a contradiction. References: [1] Ian Anderson, Combinatorics of Finite Sets , Oxford University Press, 1987. [2] Ralph P. Grimaldi, Discrete and Combinatorial Mathematics (3rd Edition), Addison–Wesley, 1994. II. Solution to part (a) by Shawn Godin, St. Joseph Scollard Hall, North Bay, Ontario. [1995 : 311] We show that the only solutions (up to permutations) are (a1 , a2 , a3 , a4 ) = ( 1, 2, 3, 5) and (2, 2, 2, 2). The given product equals A + B, where A = =
1 + (a1 a2 a3 + a1 a2 a4 + a1 a3 a4 + a2 a3 a4 )ω 3 1 + (a1 a2 a3 + a1 a2 a4 + a1 a3 a4 + a2 a3 a4 ),
which is an integer, and B
= (a1 + a2 + a3 + a4 )ω +(a1 a2 + a1 a3 + a1 a4 + a2 a3 + a2 a4 + a3 a4 )ω 2 + a1 a2 a3 a4 ω 4 = (a1 + a2 + a3 + a4 )ω +(a1 a2 + a1 a3 + a1 a4 + a2 a3 + a2 a4 + a3 a4 )ω + a1 a2 a3 a4 ω.
Now to get rid of any imaginary parts the coefficients of ω and ω must be the same. (And note that since the real parts of ω and ω are 1/2, if these coefficients are equal, B, and thus, the given product, will be an integer.) Thus, we need
−
a1 + a2 + a3 + a4 + a1 a2 a3 a4 = a1 a2 + a1 a3 + a1 a4 + a2 a3 + a2 a4 + a3 a4 . Without loss of generality assume 1
≤ a1 ≤ a2 ≤ a3 ≤ a4. If a 1a2 ≥ 6, then
(4)
58
a1 + a2 + a3 + a4 + a1 a2 a3 a4 > 6a3 a4
a1 a2 + a1 a3 + a1 a4 + a2 a3 + a2 a4 + a3 a4 ;
≥
so that a1 a2 < 6, and we have two cases: a1 = 1 and a1 = 2. Case 1: a1 = 1. Then (4) reduces to 1 + a2 a3 a4 = a2 a3 + a2 a4 + a3 a4 .
(5)
If a 2
≥ 3 then 1 + a2 a3 a4 > 3a3 a4
a2 a3 + a2 a4 + a3 a4 ;
≥
so that a2 = 1 or 2. If a2 = 1, (5) reduces to 1 = a 3 + a4 which has no positive solution. If a 2 = 2, (5) reduces to 1+ a3 a4 = 2(a3 + a4 ) which has solution a 3 = 3, a4 = 5. [Editor’s note . We borrow from Janous’s solution and write this equation as (a3 2)(a4 2 ) = 3,
−
−
which makes it clear that a 3 = 3, a 4 = 5 is the only possibility.] Case 2: a1 = 2. Since a 1 a2 < 6, a 2 = 2 also, and (4) reduces to a 3 a4 = a 3 + a4 , which has a 3 = 2, a 4 = 2 as its only positive solution. III. Generalization of part (b) by Gerd Baron, Technische Universit¨ at Wien, Austria . [1995 : 311] We will consider the following more general situation: for n, m positive integers and ω m = 1, determine all sets a1 , . . ., a n of positive integers such that ni=1 (1+ ai ω) is an integer . We will prove:
{
}
THEOREM. If m > 3 is prime and ω is a complex mth root of unity, then, for m n 2m 3 and a1 , a2 , . . ., an non-negative integers, the product n i=1 (1 + ai ω) is an integer only if some a i = 0.
≤ ≤
−
Note that for m = 5 and 5
≤ n ≤ 7, we get no solutions and (b) is done.
Proof of the theorem . Assume all a i > 0. As a polynomial in ω , n
n
(1 + ai ω) =
i=1
pk ω k =: P (ω)
k=0
where the p k are the elementary symmetric functions in a i , with p0 = 1, Let n
p1 =
ai ,
and pn =
ai .
≤ 2m − 3. Reducing P (ω) modulo the relation ω m = 1 we get m 1
Q(ω) =
−
n m
pk ω +
k=n m+1
−
k
−
( pk + pk+m )ω k .
k=0
If m is prime, then the polynomial R(ω) = ω m−1 +ω m−2 + +ω +1 is irreducible, and Q(ω) = u is an integer exactly if the polynomial Q(ω) u is an integer multiple of R(ω); that is, all coefficients of Q(ω) but the constant term are equal.
··· −
59
[Editor’s note . It is known that if m is prime, the polynomial x m−1 + xm−2 + + x + 1 is irreducible; that is, it cannotbe factored into polynomials of smaller degree with integer coefficients. For example, see exercise 19, page 84 of Ed Barbeau’s Polynomials (Springer–Verlag, 1989), or most any text on abstract algebra. Now since ω is a root of the polynomials R(ω) and Q(ω) u, it will be a root of their greatest common divisor, which must be R(ω) since R(ω) is irreducible; thus, Q(ω) u must be a multiple of R(ω), and a constant multiple since they both have degree m 1.]
···
−
−
−
If we can show that p k = p k+1 for some k with n m < k < m 1, we are done and there is no solution with all a i > 0. We claim that if m is a prime greater than 3, and m n 2m 3, then there is such a k with pk+1 pk > 0. To prove this we take a i = 1 + bi and let q 0 , q 1 , . . . , qn be the elementary symmetric functions in the b i ’s. To calculate the p k ’s in terms of the q j ’s, note that
≤ ≤
−
−
−
−
n
n
(x + ai ) =
i=1
(x + 1 + bi )
i=1
gives n
n
n k
pk x −
=
− − q j (x + 1)n−j
j =0
k=0
n j
n
=
q j
j =0
−
n
j
k
k=0
n
k
x
=
n
n n
q j
j =0
k=j
−
j n−k x ; k
therefore, [exchanging the order of summation and equating coefficients] k
pk =
j =0
Thus, the difference p k+1 rj
= =
k
− − − − − − − − − − − − −− − n n
j q j = k
k+1 j =0 r j q j ,
pk equals
n j n j = k + 1 j k j n j n 2k + j 1 k j k j + 1
−
j =0
n j q j . k j
where
n j k j
n k k + 1 j
− − 1 −
(6)
−
for j < k + 1, and rk+1 = 1. Since ai is a positive integer, it follows that bi 0 and therefore, q j 0 and q 0 = 1 > 0. If m n < 2m 3, then 0 < n m + 1 < m 2 < m 1; therefore, we can set k = n m + 1, and n m < k < m 1 will hold. Moreover 2k + 1 = 2(n m + 1 ) + 1 < n, so that, from (6) all r j are positive [in particular, r0 > 0], and therefore, p k+1 pk r0 q 0 > 0.For n = 2m 3, putting k = n m+1 = m 2 again satisfies n m < k < m 1. Also, 2k+1 = 2m 3 = n, so that, by (6) r 0 = 0 but r j > 0 for j 1. Hence, p k+1 pk > 0 unlessq j = 0 for all j 1; that is, b i = 0 for all i. But then all a i = 1, so that p k = nk for all k.
−
− ≥
−
≥
≤ −
−
−
− ≥
− ≥
−
−
−
−
−
−
≥
−
−
60
Editor’s note . Baron now computes the difference of the coefficients of ω m−3 and ω and shows it is non-zero, which finishes the proof. But it is easier to compare the coefficients of ω m−2 and ω m−3 : we get (since m 3 = n m) that these coefficients are respectively p m−2 and p m−3 + p2m−3 ; that is,
−
2m 3 m 2
− −
2m 3 2m + m 3 2m
and
− −
−3 −3
2m 3 +1; m 3
=
−
− −
however it is known (and easy to see) that consecutive binomial coefficients n and k− for n > 2 never differ by only 1, so that we are done. 1
n k
Baron then proves part (a) separately (finding both solutions), since his theorem does not apply to m = 3. He also states that if we allow some of the a i ’s to be zero,there are two more solutions to (a), namely (0 , 1, 1, 1) and (0, 0, 2, 2).
1362. [1988 : 202] Determine the sum n
n
n j + k
n j
n −j −2k ω , k
j =0 k=0
where w is a primitive cube root of unity. Solution by G.P. Henderson, Campbellcroft, Ontario. [1989 : 249] The coefficient of tn in (1 + xt)a (1 + yt)b (1 + zt)c is
a j
b k
j +k n
≤
n
c j
− −k
xi y k z n−j−k .
Setting a = b = c = n, x = ω −1 , y = ω −2 , z = 1, we see that the required sum is the coefficient of t n in t (1 + t) 1 + ω
Therefore, the sum is
n n/3
t 1+ 2 ω
n
= [(1 + t)(ω + t)(ω 2 + t)]n = (1 + t2 )n .
or 0 according as n is or is not divisible by 3.
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