Tạp chí toán học AMM của Mỹ
Trang 1PROBLEMS AND SOLUTIONS
Edited by Gerald A Edgar, Doug Hensley, Douglas B West
with the collaboration of Paul T Bateman, Mario Benedicty, Itshak Borosh, Paul Bracken, Ezra A Brown, Randall Dougherty, Tam´as Erd´elyi, Zachary Franco, Chris-tian Friesen, Ira M Gessel, Jerrold Grossman, Frederick W Luttmann, Vania Mas-cioni, Frank B Miles, Richard Pfiefer, Cecil C Rousseau, Leonard Smiley, John Henry Steelman, Kenneth Stolarsky, Richard Stong, Walter Stromquist, Daniel Ull-man, Charles Vanden Eynden, and Fuzhen Zhang
Proposed problems and solutions should be sent in duplicate to theMONTHLY
problems address on the inside front cover Submitted solutions should arrive at that address before May 31, 2008 Additional information, such as generaliza-tions and references, is welcome The problem number and the solver’s name and address should appear on each solution An asterisk (*) after the number of
a problem or a part of a problem indicates that no solution is currently available.
PROBLEMS
11334 Proposed by Jonathan Bober and Jeffrey Lagarias, University of Michigan,
Ann Arbor, MI Let x and y be positive integers with x2− 3y2 = 1
(a) Show that if x cannot be written as 2 n, 2n± 1, 3 · 2n, or 3· 2n± 1 for any natural
number n, then x y has more than 3 distinct prime factors.
(b) Show that if x = 2n − 1 with n > 3, then y has at least three distinct prime factors.
11335 Proposed by Juan L´opez Gonz´alez, Madrid, Spain Let σ (n) denote the sum of
the divisors of n.
(a) Find, with proof, the least positive integer m such that σ (6m) < σ (6m + 1).
(b) Show that there are infinitely many m such that σ (6m) < σ (6m + 1).
(cf G Martin, The Smallest Solution ofφ(30n + 1) < φ(30n) Is , this MONTHLY
106 (1999) 449-451.)
11336 Proposed by Donald Knuth, Stanford University, Stanford, CA A
near-deBruijn cycle of order d is a cyclic sequence of 2 d − 1 zeros and ones in which all 2d − 1 substrings of length d are distinct For all d > 0, construct a near-deBruijn cycle of order d+ 1 such that the front and back substrings of length 2d − 1 are both
near-deBruijn cycles of order d (Thus, for example, 1100010 is near-deBruijn of
order 3, while 110 and 010 are both near-deBruijn of order 2.)
11337 Proposed by Marian Tetiva, National College “Gheorghe Ros¸ca Codreanu”,
Bˆırlad, Romania Suppose in triangle A BC we have opposite sides of lengths a, b, and
c, respectively, with a ≤ b ≤ c Let w a andw b be the lengths of the bisectors of the
angles A and B respectively Show that a + w a ≤ b + w b
11338 Proposed by Ovidiu Furdui, Cluj, Romania Let denote the classical gamma
function, and let G (n) =n
k=1(1/k) Find
lim
G (n + 1)1/(n+1) − G(n)1/n
.
Trang 211339 Proposed by Jos´e Luis D´ıaz-Barrero, Universitat Polit`ecnica de Catalunya,
Barcelona, Spain Let F n and L n denote the nth Fibonacci and Lucas numbers, respec-tively Prove that for all n ≥ 1,
1 2
F1/F n
n
≤ 2 − F n+1
F 2n
(The Fibonacci and Lucas numbers are given by the recurrence a n+1 = a n + a n−1, with
F0 = 0, F1= 1, L0 = 2 and L1= 1.)
11340 Proposed by ´ Oscar Ciaurri and Luz Roncal, Universidad de la Rioja, Logro˜no, Spain An umbrella of radius 1 meter is spun with angular velocity ρ in the xz-plane
about an axis (call it the y-axis) parallel to the ground It is wet, and drops of water
crawl along the ribs and fly off as they reach their ends
Each drop leaves the umbrella with a velocity vector equal to the velocity of the tip of the rib at the point where it exited It then follows a parabolic trajectory If a drop spins off while on the downspin, then the high point in its arc will be the point
of departure Otherwise, the high point is the vertex of a parabolic arc in the x z-plane Determine a parameterized family P ρof polynomials in two variables such that when-everρ2 > g, the various arc vertices reached by the water droplets all lie on the curve
P ρ (x, z) = 0 (Here g denotes the magnitude of the downward acceleration due to
gravity.)
The figure shows the case ρ2/g = 2,
with the umbrella spinning
counterclock-wise
SOLUTIONS
A Condition on the Commutator
11196 [2006, 79] Proposed by Mohammad Hossein Mehrabi, Iran University of
Sci-ence and Technology, Tehran, Iran Let A and B be real n × n matrices Show that if
A B − B A is invertible and A2+ B2=√3(AB − B A), then n is a multiple of 6 Solution by P´al P´eter D´alyay, De´ak Ferenc High School, Szeged, Hungary From (A + i B)(A − i B) = A2+ B2− i(AB − B A) = (√3− i)(AB − B A),
it follows that det[(A + i B)(A − i B)] = (√3− i) ndet(AB − B A) Since AB − B A
is invertible, so is(A + i B)(A − i B) We conclude that
det[(A + i B)(A − i B)] = det(A + i B) det(A − i B) = det(A + i B)det(A + i B) > 0 Since A and B are real, also det (AB − B A) is real; hence (√3− i) n is real Since
√
3− i = 2e −πi/6 , we conclude that n is a multiple of 6.
Also solved by S Amghibech (Canada), R Chapman (U K.), Y Dumont (France), O Furdui, J.-P Gri-vaux (France), C C Heckman, E A Herman, O P Lossers (Netherlands), M Omarjee (France), S Pierce,
A K Shafie (Iran), R Stong, M Tetiva (Romania), J Vinuesa (Spain), L Zhou, NSA Problems Group, and the proposer.
Trang 3Can One Recover a (Spherical) Triangle from Its Medial Triangle?
11201 [2006, 179] Proposed by Robert Russell, New York, NY Given the midpoints
of the sides of a spherical triangle, provide a construction for the original triangle.
Composite solution by Christoph Soland (Switzerland), William Dickenson, and the editors There is one configuration of midpoints, namely, mutually perpendicular, for
which many choices of original triangle work Otherwise, the underlying triangle may
be recovered by a construction
We take it to be part of the definition of a spherical triangle that the three vertices
do not lie on a common great circle We also assume (by definition) that the sides of
a spherical triangle are less than π radians Thus, to construct a spherical triangle, it
suffices to construct the three vertices LetObe the center of our sphereS Without loss of generality, we takeOto be the origin ofR3and the radius ofSto be 1 When
A and B are not antipodal, let m (A, B) be the midpoint of the spherical line segment
A B, defined as the shorter of the two portions of the great circle through A and B
delimited by A and B.
We first consider the exceptional case If any side of the medial triangle M1M2M3
isπ/2, then, we claim, all of them are, and there are infinitely many triangles with the
same three midpoints, so we cannot uniquely construct the “original triangle” For the
proof of this claim, suppose that M1⊥ M2 Without loss of generality, M1= (1, 0, 0) and M2= (0, 1, 0) By assumption, there exist distinct S1, S2, and S3, no two antipodal, such that for any permutation(i, j, k) of (1, 2, 3), m(S i , S j ) = M k Note that the ends
of a line segment in spherical geometry are nearer the midpoint than its antipode
Now if S2= (x2, y2, z2), then x2 > 0, S3= (x2, −y2, −z2) so that y2< 0, and S1 =
(−x2, −y2, z2) Now z2 = 0 since S1 = −S2, and it follows that M3= (0, 0, sign(z2)).
Thus all three axes are perpendicular Glancing over the foregoing algebra, we see that
any choice of S2= (x2, y2, z2) with x2, z2> 0 > y2and x2
2+ y2
2+ z2
2= 1 gives rise to
S2, S3, and S1for which(1, 0, 0), (0, 1, 0), and (0, 0, 1) are the vertices of the medial
triangle
We now consider the composition R = H3◦ H2◦ H1, where H j is the bijection of
Sobtained by reflecting it through the axis of±M j Either R is the identity map on S,
or it is a nontrivial rotation that fixes two antipodal points, call them±S We claim that the M j are mutually perpendicular if and only if R = I In one direction this is trivial
by coordinate algebra after setting M1 = (1, 0, 0) and so on In the other direction, suppose R = I Then for any permutation (i, j, k) of (1, 2, 3), H i ◦ H j = H k Thus
the M j are distinct, and no two are antipodal Now let P = H2M1 and Q = H3M1 If
M1 ⊥ M2, then M1and M2determine a line L on S , P, M1, and M2are distinct points
on L, and P is not antipodal to M1
Furthermore, H1Q = P since H1H3H2 = I Thus Q = H1P Since P lies on L
and H1 carries any line through M1 onto itself, Q is on L From Q = H3M1, we
have also that Q is on the line L through M1 and M3 Thus Q = M1, Q = −M1, or
L = L But Q = M1 because M3 = ±M1, and Q = −M1because that would make
P = H1Q = −M1, a contradiction Finally, if L = L , each H j transposes the poles
ofS if we take L as the equator, and so R transposes those poles, contrary to the the assumption that R = I This proves that M1 ⊥ M2, and by the same logic, all three
M jare mutually perpendicular
If, on the other hand, no two of the M j are perpendicular, then the underlying S j
may be recovered from the M j by a construction
Henceforth, we assume that no two of the M j are orthogonal Thus, R = I We
first show that the axis±S of rotation for R is not perpendicular to M1 To see this,
suppose to the contrary that H3H2H1S = S, and S ⊥ M1 Take S = H1S and S =
Trang 4H2S Then H3S = H3H2H1S = S If S and S are antipodal, then S = H2S =
−H2S Since H3S = S, M3 S, and thus M3 ⊥ M1, a contradiction If S and S are not antipodal, then, viewing these points as vectors inR3, from H2S = S it follows
that M2 (S + S ) If also S and S are not antipodal then M
3 S + S = S − S .
But(S + S ) ⊥ (S − S ) so M2⊥ M3 If instead, S and S are antipodal, then S =
S , But then H2S = S so that M
2 ±S From M1 ⊥ S we have M1 ⊥ M2 This
shows that S ⊥ M1 Now take S2 as the nearer of±S to M1 Then take S3 = H1S2,
and S1= H2S3
It remains to discuss how S2 may be located via a construction This can depend
on what tools are permitted Here, we assume that lines through points, intersections
of lines, midpoints of line segments, and perpendicular bisectors of line segments are constructible, that we can pick points on lines that are distinct from given points or their antipodes, that we can construct reflections of a point about a given axis, that
we can determine which of two points is nearer a given point if the points are not equidistant, and we can pick points not on specified lines
Consider, then, an arbitrary pair(A, B) of distinct, non-antipodal points If either of
these is fixed by R, then we have ±S2and we are done Otherwise, S2lies on the per-pendicular bisector of(A, R(A)), as well as the perpendicular bisector of (B, R(B)).
If these are distinct, their intersection again gives S2 If not, then the axis of rotation
for R lies on the intersection of the lines A B and R A R B, and that gives S2
Editorial comment This problem would be trivial in Euclidean geometry: draw
through each midpoint a line parallel to the segment joining the other two midpoints; these lines define the original triangle In spherical geometry there are no parallels, so this construction doesn’t make sense There are other snags, and the somewhat narrow definition used here for the term ‘spherical triangle’ was the only means known to the editors to avoid these
Also solved by R Stong, A Tissier (France), and the Microsoft Research Problems Group.
Differentiable and Discontinuous Densely
11221 [2006, 367] Proposed by Paolo Perfetti, University “Tor Vergata”, Rome, Italy.
Give an example of a function g from R into R such that g is differentiable everywhere,
g is differentiable on one dense subset ofR, and g is discontinuous on another dense
subset ofR
Solution by Richard Bagby, New Mexico State University, Las Cruces, NM Define
f : R → R by f (x) = x2sin2(π/x) for 0 < x < 1 and f (x) = 0 otherwise Let {q k}
be an enumeration of the rationals We will show that
g (x) =∞
k=1
2−3k f
2k (x − q k )
has the required properties
Note that f is differentiable everywhere: 0 ≤ f (x) ≤ x2shows f (0) = 0 and 0 ≤
f (x) ≤ x2sin2(π − π/x) ≤ π2(x − 1)2 shows f (1) = 0 Also note that | f (x)| is
bounded and f is discontinuous at 0
Since f is continuous and the series for g (x) converges uniformly, the sum g(x)
exists and defines a continuous function The derived series
∞
k=1
2−2k f
Trang 5converges uniformly, so g is differentiable and g (x) is the sum of this derived series.
We claim that g is discontinuous at each rational q j Write (1) as
g (x) = 2 −2 j f
2j (x − q j )+
k = j
2−2k f
2k (x − q k ).
The last series defines a function continuous at q j since each term is continuous there
and the series converges uniformly The discontinuity of f at 0 then shows that g is
discontinuous at q j
For n a positive integer, let E n =∞
k =n (q k− 21−k , q k+ 21−k ) and E =∞
Thus E n has Lebesgue measure at most 23−n and E has measure zero, so the set of
irrationals in the complement of E is dense We claim g is differentiable at every irrational x0 ∈ E Since x0is neither q k nor q k+ 2−k for any k, each term 2 −3k f (2 k (x −
q k )) is twice differentiable at x0 Let n be such that x0 ∈ E n Then the derived series
(1) reduces to a finite series, and its derivative at x0is
n−1
k=1
2−k f
To prove this finite sum is g (x0), let
α(x) =∞
2−2k f
2k (x − q k ).
The term f
2k (x − q k )vanishes when|x − x0| ≤ 2−k, so
α( x ) − α(x0) ≤C
2−k ≤|x−x0 |
2−k |x − x0| ≤ 2C|x − x0|2,
where C is a bound for | f | Hence α (x0) = 0, so g (x0) is indeed given by the finite
series (2)
Also solved by J H Lindsey II, M D Meyerson, A Stadler (Switzerland), R Stong, Szeged Problem Solving Group “Fej´ental´altuka” (Hungary), GCHQ Problem Solving Group (U K.), and the proposer.
Snapshots for Velocities
11223 [2006, 459] Proposed by Christopher Hillar, Texas A& M University, College
Station, TX, and Lionel Levine, University of California at Berkeley, Berkeley, CA.
Consider n unlabeled particles moving each at its own constant velocity along the real line An observer is promised some number P of snapshots of the particles, to be taken
at uniformly spaced intervals of time When particles coincide, the snapshot will show how many are at a given point
(a) Show that if P = n + 1 then the observer can determine the velocity of each of the
particles
(b∗) As a function of n, what is the minimum value of P that will suffice to ensure that the observer can determine all n velocities?
Solution to (a) by S C Locke, Florida Atlantic University, Boca Raton, FL Suppose
the kth snapshot is taken at time t k and the positions of the n particles at that time
are(x k , j ) n
j=1, where x k , j is not necessarily the position of the j th particle Let G =
{(t k , x k , j ): 1 ≤ k ≤ n + 1, 1 ≤ j ≤ n} If p j (t) is the position of the jth particle at
time t, then the points L j = {(t, p j (t)): t ∈ R} would constitute a line meeting G in
exactly n+ 1 points (not counted with multiplicity)
Trang 6If a line L meets G in at least n + 1 points (and therefore exactly n + 1 points), then at least two of these points are generated by the same particle, say particle j Thus L = L j , since two distinct points determine a line The lines meeting G in n+ 1 points are therefore exactly the lines representing the motion of the individual particles, and the velocities of the particles are the slopes of these lines
Editorial comment No solutions to (b) were received Several solvers noted that
for n ≤ 3, n + 1 snapshots are required For n = 3, the three snapshots (−2, 0, 2),
(−1, 0, 1) and (−2, 0, 2) are ambiguous The GCHQ Problem Solving Group verified
that for n= 4, four snapshots suffice The best lower bound was due to Petr Skovron
Let P (n) be the minimum number of snapshots required for n particles Suppose the
n trajectories { f i (t)} n
i=1 are indistinguishable from {g i (t)} n
i=1 with snapshots at times
1, , P − 1 For any constant a, the 2n trajectories { f i (t) + a(t − P)} n
i=1∪ {g i (t) −
a (t − P)} n
i=1are indistinguishable from{ f i (t) − a(t − P)} n
i=1∪ {g i (t) + a(t − P)} n
i=1
with snapshots at times 1, , P Hence P(2n) ≥ P(n) + 1 It follows that P(n) ≥
log2n + 2
Part (a) also solved by D Beckwith, K Bernstein, D R Bridges, P Budney, R Chapman (U K.), J H Lindsey II, L Pebody, R E Prather, F Yang, GCHQ Problem Solving Group (U K.), Houghton College Problem Solving Group, Szeged Problem Solving Group “Fej´ental´altuka” (Hungary), and the proposer.
A Unique Solution
11226 [2006, 460] Proposed by Franck Beaucoup, Ottawa, Canada, and Tam´as
Erd´elyi, Texas A& M University, College Station, TX Let a1, , a nbe real numbers,
each greater than 1 For n ≥ 2, show that there is exactly one solution in the interval
(0, 1) to
n
j=1
(1 − x a j ) = 1 − x.
Solution by Microsoft Research Problems Group, Redmond, WA If a ≥ 1 and x ≥ 0, then x a ≥ 1 + a(x − 1) by the mean value theorem (MVT) Equality occurs only for
a = 1 or x = 1 If f (x) =n
j=1(1 − x a j ) for 0 ≤ x ≤ 1, then
0≤ f (x) ≤ (1 − x a1)(1 − x a2) ≤ a1a2(1 − x)2.
Hence f (x) < 1 − x as x approaches 1− On the other hand, 1≥ f (x) ≥ 1 − (x a1+
· · · + x a n ) shows that f (x) > 1 − x as x approaches 0+ By the intermediate value
theorem, f (x) = 1 − x has a root in (0, 1).
There is only one such root Let g (x) = log(1 − x) − log f (x) Suppose g(0) =
g (x1) = g(x2) = 0 where 0 < x1 < x2 < 1 By MVT, h(x) = (1 − x)g (x) has
at least one zero in (0, x1) and another in (x1, x2) By MVT again, h has a root
in (0, x2) However, h(x) = −1 + n
j=1a j x a j−1(1 − x)(1 − x a j )−1, so h (x) = n
j=1a j x a j−2(x a j − 1 − a j x + a j )(1 − x a j )−2and is thus positive on(0, 1).
Also solved by K F Andersen (Canada), R Bagby, P P D´alyay (Hungary), J H Lindsey II, O P Lossers (Netherlands), T L McCoy (Taiwan), R Mortini (France), L Pebody, J Rooin and A Mahmoodi (Iran), A Stadler (Switzerland), R Stong, BSI Problems Group (Germany), GCHQ Problem Solving Group (U K.), and the proposers.
Trang 7A Triangle Inequality
11228 [2006, 460] Proposed by Marian Tetiva, Bˆırlad, Romania Prove that in an
acute triangle with angles A, B, and C radians,
(1 − cos A)(1 − cos B)(1 − cos C)
cos A cos B cos C
27(tan A + tan B)(tan A + tan C)(tan B + tan C) . Solution by Minh Can, Irvine Valley College, Irvine, CA.
(tan A + tan B + tan C)3cos A cos B cos C
(tan A + tan B)(tan B + tan C)(tan C + tan A)(1 − cos A)(1 − cos B)(1 − cos C)
sin C
cos A cos B
sin A cos B cos C
sin B cos C cos A (1 − cos A)(1 − cos B)(1 − cos C)
= (sin A)2(sin B)2(sin C)2
(1 − cos A)(1 − cos B)(1 − cos C) =
(1 − cos2A )(1 − cos2B )(1 − cos2C ) (1 − cos A)(1 − cos B)(1 − cos C)
= (1 + cos A)(1 + cos B)(1 + cos C) ≤
3+ cos A + cos B + cos C
3
3
≤ 3+ 3 cos
(A + B + C)/3
3
3
= 27
8 .
The last inequality follows from the concavity of cos x on the interval [0, π/2] The
inequality before that follows from the arithmetic–geometric mean inequality Equality holds if and only if the triangle is equilateral
Also solved by S Amghibech (Canada), A Arkady, A R Avidon, M Battaille (France), D Beckwith, A Bun-gale (India), R Chapman (U K.), G H Chung, P P D´alyay (Hungary), P De (Ireland), O Faynshteyn (Germany), D Fleischman, M Goldenberg, M Hajja (Jordan), E A Herman, Y.-J Kuo, J Lee (Korea),
O P Lossers (Netherlands), M Mabuchi (Japan), D J Moore, L Pebody, D Perkins, C R Pranasachar (In-dia), J Rooin (Iran), V Schindler (Germany), S Shaebani (Iran), A Stadler (Switzerland), R Stong, T Tam,
S Varosanec (Croatia), J Vinuesa (Spain), M Vowe (Switzerland), GCHQ Problem Solving Group (U K.), Microsoft Research Problems Group, Szeged Problem Solving Group “Fej´ental´altuka” (Hungary), and the proposer.
A Function Inequality
11232 [2006, 567] Proposed by Michael W Botsko, Saint Vincent College, Latrobe,
PA Let f be a continuous mapping fromR into R that is bounded below Show that
there exists a real number x0 such that f (x0) − f (x) < |x − x0| holds for all x other than x0
Solution by Albert Stadler, D¨ubendorf, Switzerland Put g (x) := f (x) + |x|/2 Then
g is continuous, bounded below, and g (x) → +∞ as x → ±∞ Therefore there is a
real number x0such that g (x0) = min x∈Rg (x) Thus f (x) + |x|/2 ≥ f (x0) + |x0|/2
or f (x0) − f (x) + |x0|/2 − |x|/2 ≤ 0 We conclude that for x = x0,
f (x0) − f (x) − |x − x0| < f (x0) − f (x) − |x − x0|/2
≤ f (x0) − f (x) + |x0|/2 − |x|/2 ≤ 0.
Trang 8Also solved by 41 others, and the proposer.
A Derivative Formula
11233 [2006, 568] Proposed by Robert Downes, Mountain Lakes High School,
Moun-tain Lakes, NJ Show that for positive integer n, and for x = 0,
d n
d x n
x n−1sin
1
x
= (−1) n
x n+1 sin
1
x + n π 2
.
Solution by Jean-Pierre Grivaux, Paris, France We will show more generally that if
f is an n-times differentiable real-valued function, then
d n
d x n
x n−1f
1
x
= (−1) n
x n+1 f (n)
1
x
.
The proof is by induction on n The case n = 1 is obvious Suppose it is true for
n − 1 Applying the inductive hypothesis, and taking g(x) = x n−2f (1/x) in Leibniz’s
formula
d n
d x n (xg(x)) = x d n g
d x n + nx d n−1g
d x n−1,
we compute
d n
d x n
x n−1f
1
x
= x d n
d x n
x n−2f
1
x
+ nx d n−1
d x n−1
x n−2f
1
x
= x d
d x
(−1) n−1
1
x
+ nx
(−1) n−1
1
x
= x
(−1) n n
1
x
+(−1) n
x n+2 f (n)
1
x
+ (−1) n−1n
1
x
= (−1) n
x n+1 f (n)
1
x
.
Specializing to f (x) = sin(x) gives the desired result.
Editorial comment The more or less equivalent result (obtained by setting f (x) = e x,)
d n
d x n
x n−1e1/x
= (−1) n
x −n−1 e1/x ,
appears as a problem in 1850 Exercices de Math´ematiques Pos´es `a l’Oral du CAPES de
Math´ematiques et des Concours des Grande ´ Ecoles, by L Moisotte The most general
formula of this kind submitted was given by O P Lossers, who showed
d k
d x k
x m f
1
x
=
k
j=0
k!
j!
m − j
k − j
x m −k− j (−1) j
f ( j)
1
x
,
from which the above formula follows by taking k = n and m = n − 1.
Also solved by 74 other readers and the proposer.
...(1 − cos A) (1 − cos B) (1 − cos C) =
(1 − cos2A ) (1 − cos2B ) (1 − cos2C ) (1 − cos A) (1 − cos B) (1 −... C)(tan C + tan A) (1 − cos A) (1 − cos B) (1 − cos C)
sin C
cos A cos B
sin A cos B cos C
sin B cos C cos A (1 − cos A) (1 − cos B) (1 − cos C)... ) for ≤ x ≤ 1, then
0≤ f (x) ≤ (1 − x a1< /sub>) (1 − x a2) ≤ a1< /small>a2 (1 − x)2.