Tạp chí toán học AMM tháng 1 năm 2008

PROBLEMS 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 the MONTHLY
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 x
2
− 3y
2
= 1.
(a) Show that if x cannot be written as 2
n
,2
n
± 1, 3 · 2
n

,or3· 2
n
± 1 for any natural
number n,thenxy has more than 3 distinct prime factors.
(b) Show that if x = 2
n
− 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 of φ(30n + 1)<φ(30n) Is ,thisM
ONTHLY
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 2
d
− 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 2
d
− 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”,
B
ˆ
ırlad, Romania. Suppose in triangle ABC we have opposite sides of lengths a, b,and
c, respectively, with a ≤ b ≤ c.Letw
a
and w
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
n→∞

G(n + 1)
1/(n+1)
− G(n)
1/n

.
January 2008]

PROBLEMS AND SOLUTIONS 71
11339. Proposed by Jos
´
eLuisD
´
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

F
1/F
n
n
+ L
1/L
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
F
0
= 0, F
1
= 1, L
0
= 2andL
1
= 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 xz-plane.

Determine a parameterized family P
ρ
of polynomials in two variables such that when-
ever ρ
2
> 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 ρ
2
/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
AB − BAis invertible and A
2
+ B
2
=
√
3(AB − BA),thenn is a multiple of 6.
Solution by P
´
al P
´

eter D
´
alyay, De
´
ak Ferenc High School, Szeged, Hungary. From
(A + iB)( A − iB) = A
2
+ B
2
− i(AB − BA) = (
√
3 −i)(AB − BA),
it follows that det[(A + iB)(A −iB)]=(
√
3 − i)
n
det(AB − BA).SinceAB − BA
is invertible, so is (A + iB)(A −iB). We conclude that
det[(A +iB)(A −iB)]=det(A +iB) det(A −iB) = det(A +iB)
det(A +iB)>0.
Since A and B are real, also det(AB − BA) 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.
72
c
 THE MATHEMATICAL ASSOCIATION OF AMERICA [Monthly 115
Can 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. Let
O be the center of our sphere S. Without
loss of generality, we take
O to be the origin of R
3
and the radius of S to be 1. When
A and B are not antipodal, let m(A, B) be the midpoint of the spherical line segment
AB, 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 M
1
M
2
M

3
is π/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 M
1
⊥ M
2
. Without loss of generality, M
1
= (1, 0, 0)
and M
2
= (0, 1, 0). By assumption, there exist distinct S
1
, S
2
,andS
3
, 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 S
2
= (x

2
, y
2
, z
2
),thenx
2
> 0, S
3
= (x
2
, −y
2
, −z
2
) so that y
2
< 0, and S
1
=
(−x
2
, −y
2
, z
2
).Nowz
2
= 0sinceS
1

=−S
2
, and it follows that M
3
= (0, 0, sign(z
2
)).
Thus all three axes are perpendicular. Glancing over the foregoing algebra, we see that
any choice of S
2
= (x
2
, y
2
, z
2
) with x
2
, z
2
> 0 > y
2
and x
2
2
+ y
2
2
+ z
2

2
= 1givesriseto
S
2
, S
3
,andS
1
for which (1, 0, 0), (0, 1, 0),and(0, 0, 1) are the vertices of the medial
triangle.
We now consider the composition R = H
3
◦ H
2
◦ H
1
,whereH
j
is the bijection of
S obtained 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 M
1
= (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 = H
2
M
1
and Q = H
3
M
1
.If
M
1
⊥ M
2
,thenM
1
and M
2
determine a line L on S, P, M
1
,andM
2
are distinct points

on L,andP is not antipodal to M
1
.
Furthermore, H
1
Q = P since H
1
H
3
H
2
= I . Thus Q = H
1
P.SinceP lies on L
and H
1
carries any line through M
1
onto itself, Q is on L.FromQ = H
3
M
1
,we
have also that Q is on the line L

through M
1
and M
3
. Thus Q = M

1
, Q =−M
1
,or
L = L

.ButQ = M
1
because M
3
=±M
1
,andQ =−M
1
because that would make
P = H
1
Q =−M
1
, a contradiction. Finally, if L = L

, each H
j
transposes the poles
of
S if we take L as the equator, and so R transposes those poles, contrary to the the
assumption that R = I . This proves that M
1
⊥ M
2

, and by the same logic, all three
M
j
are 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 M
1
. To see this,
suppose to the contrary that H
3
H
2
H
1
S = S,andS ⊥ M
1
.TakeS

= H
1
S and S


=
January 2008]
PROBLEMS AND SOLUTIONS 73
H
2
S

.ThenH
3
S

= H
3
H
2
H
1
S = S.IfS

and S

are antipodal, then S

= H
2
S

=
−H

2
S.SinceH
3
S

= S, M
3
 S, and thus M
3
⊥ M
1
, a contradiction. If S

and S

are
not antipodal, then, viewing these points as vectors in R
3
, from H
2
S

= S

it follows
that M
2
 (S

+ S


).IfalsoS

and S are not antipodal then M
3
 S

+ S = S

− S

.
But (S

+ S

) ⊥ (S

− S

) so M
2
⊥ M
3
. If instead, S

and S are antipodal, then S

=
S


,ButthenH
2
S

= S

so that M
2
±S.FromM
1
⊥ S we have M
1
⊥ M
2
.This
shows that S ⊥ M
1
.NowtakeS
2
as the nearer of ±S to M
1
.ThentakeS
3
= H
1
S
2
,
and S

1
= H
2
S
3
.
It remains to discuss how S
2
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,thenwehave±S
2
and we are done. Otherwise, S
2
lies 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 S
2
. If not, then the axis of rotation
for R lies on the intersection of the lines AB and RA RB, and that gives S
2
.
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 of R,andg

is discontinuous on another dense
subset of R.
Solution by Richard Bagby, New Mexico State University, Las Cruces, NM. Define
f : R → R by f (x) = x
2
sin
2
(π/x) for 0 < x < 1and f (x) = 0 otherwise. Let {q
k
}
be an enumeration of the rationals. We will show that
g(x) =
∞

k=1
2
−3k
f


2
k
(x − q
k
)

has the required properties.
Note that f is differentiable everywhere: 0 ≤ f (x) ≤ x
2
shows f

(0) = 0and0≤
f (x) ≤ x
2
sin
2
(π − π/x) ≤ π
2
(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


2
k
(x − q
k
)

(1)
74
c

THE MATHEMATICAL ASSOCIATION OF AMERICA [Monthly 115
converges 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


2
j
(x − q
j
)

+

k=j
2
−2k
f


2
k
(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


at0thenshowsthatg

is
discontinuous at q
j
.
For n a positive integer, let E
n
=

∞
k=n
(q
k
− 2
1−k
, q
k
+ 2
1−k
) and E =

∞
n=1
E
n
.
Thus E
n

has Lebesgue measure at most 2
3−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 x
0
∈ E.Sincex
0
is 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 x
0
.Letn be such that x
0
∈ E
n
. Then the derived series
(1) reduces to a finite series, and its derivative at x
0

is
n−1

k=1
2
−k
f


2
k
(x
0
− q
k
)

. (2)
To prove this finite sum is g

(x
0
),let
α(x) =
∞

k=n
2
−2k
f



2
k
(x − q
k
)

.
The term f


2
k
(x − q
k
)

vanishes when |x − x
0
|≤2
−k
,so


α(x) − α(x
0
)



≤ C

2
−k
≤|x −x
0
|
2
−k
|x − x
0
|≤2C|x − x
0
|
2
,
where C is a bound for | f

|. Hence α

(x
0
) = 0, so g

(x
0
) 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
,wherex
k, j
is not necessarily the position of the jth particle. Let G =
{(t

k
, x
k, j
) : 1 ≤ k ≤ n + 1, 1 ≤ j ≤ n}.Ifp
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).
January 2008]
PROBLEMS AND SOLUTIONS 75
If 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,the2n trajectories { f
i
(t) + a(t − P)}
n
i=1
∪{g
i
(t) −
a(t − P)}
n
i=1
are 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) ≥
log

2
n+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 a
1
, ,a
n
be 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 ≥ 1andx ≥ 0,
then x
a
≥ 1 +a(x − 1) by the mean value theorem (MVT). Equality occurs only for
a = 1orx = 1. If f (x) =

n
j=1
(1 − x
a
j
) for 0 ≤ x ≤ 1, then
0 ≤ f (x) ≤ (1 − x
a
1
)(1 − x
a
2
) ≤ a
1
a
2
(1 − x)
2
.
Hence f (x)<1 − x as x approaches 1
−
. On the other hand, 1 ≥ f (x) ≥ 1 − (x
a

1
+
···+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(x
1
) = g(x
2
) = 0where0< x
1
< x
2
< 1. By MVT, h(x) = (1 − x)g

(x) has
at least one zero in (0, x
1
) and another in (x
1
, x
2
). By MVT again, h

has a root

in (0, x
2
).However,h(x) =−1 +

n
j=1
a
j
x
a
j
−1
(1 − x)(1 − x
a
j
)
−1
,soh

(x) =

n
j=1
a
j
x
a
j
−2
(x

a
j
− 1 − a
j
x + a
j
)(1 − x
a
j
)
−2
and 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.
76
c
 THE MATHEMATICAL ASSOCIATION OF AMERICA [Monthly 115
A Triangle Inequality
11228 [2006, 460]. Proposed by Marian Tetiva, B
ˆ
ırlad, Romania. Prove that in an
acute triangle with angles A, B,andC radians,
(1 − cos A)(1 − cos B)(1 − cos C)
cos A cos B cos C
≥
8(tan A + tan B + tan C)

3
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)
3
cos 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)
=
(tan A + tan B + tan C)
3
cos A cos B 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 − cos
2

A)(1 − cos
2
B)(1 − cos
2
C)
(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 + 3cos

(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 from R into R that is bounded below. Show that
there exists a real number x
0
such that f (x
0
) − f (x)<|x − x
0
| holds for all x other
than x
0
.
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 x

0
such that g(x
0
) = min
x∈R
g(x). Thus f (x) +|x|/2 ≥ f (x
0
) +|x
0
|/2
or f (x
0
) − f (x) +|x
0
|/2 −|x|/2 ≤ 0. We conclude that for x = x
0
,
f (x
0
) − f (x) −|x − x
0
| < f (x
0
) − f (x) −|x − x
0
|/2
≤ f (x
0
) − f (x) +|x
0

|/2 −|x|/2 ≤ 0.
January 2008]
PROBLEMS AND SOLUTIONS 77
Also 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,andforx = 0,
d
n
dx
n

x
n−1
sin

1
x

=
(−1)
n
x
n+1
sin

1
x
+
nπ

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
dx
n

x
n−1
f

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−2
f (1/x) in Leibniz’s
formula
d
n
dx
n
(xg(x)) = x
d
n
g
dx
n
+ nx
d
n−1
g
dx
n−1
,
we compute
d
n
dx
n

x
n−1
f


1
x

= x
d
n
dx
n

x
n−2
f

1
x

+ nx
d
n−1
dx
n−1

x
n−2
f

1
x


= x
d
dx

(−1)
n−1
x
n
f
(n−1)

1
x

+ nx

(−1)
n−1
x
n
f
(n−1)

1
x

= x

(−1)
n

n
x
n+1
f
(n−1)

1
x

+
(−1)
n
x
n+2
f
(n)

1
x

+
(−1)
n−1
n
x
n−1
f
(n−1)

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
dx
n

x
n−1
e
1/x

= (−1)
n

x
−n−1
e
1/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
dx
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.
78
c
 THE MATHEMATICAL ASSOCIATION OF AMERICA [Monthly 115