amer.math.monthly.124.2.179.pdf

Problems and Solutions
Source: The American Mathematical Monthly, Vol. 124, No. 2 (February 2017), pp. 179-187
Published by: Mathematical Association of America
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PROBLEMS AND SOLUTIONS
Edited by Gerald A. Edgar, Daniel H. Ullman, Douglas B. West
with the collaboration of Paul Bracken, Ezra A. Brown, Daniel Cranston, Zachary Franco,
Christian Friesen, László Lipták, Rick Luttmann, Frank B. Miles, Leonard Smiley, Kenneth
Stolarsky, Richard Stong, Walter Stromquist, Daniel Velleman, and Fuzhen Zhang.
Proposed problems should be submitted online at
http: // www. americanmathematicalmonthly. submittable. com/ submit.
Proposed solutions to the problems below should be submitted on or before June
30, 2017 via the same link. More detailed instructions are available online. Proposed
problems must not be under consideration concurrently at any other journal nor be
posted to the internet before the deadline date for solutions. An asterisk (*) after the
number of a problem or a part of a problem indicates that no solution is currently
available.
PROBLEMS
11957. Proposed by Éric Pité, Paris, France. Let m and n be two integers with n ≥ m ≥ 2.
Let S(n, m) be the Stirling number of the second kind, i.e., the number of ways to partition
a set of n objects into m nonempty subsets. Show that
nm
S(n, m) ≥ mn

n
m

.
11958. Proposed by Kent Holing, Trondheim, Norway.
(a) Find a condition on the side lengths a, b, and c of a triangle that holds if and only if the
nine-point center lies on the circumcircle.
(b) Characterize the triangles whose nine-point center lies on the circumcircle and whose
incenter lies on the Euler line.
11959. Proposed by Donald Knuth, Stanford University, Stanford, CA. Prove that, for any
n-by-n matrix A with (i, j)-entry ai, j and any t1, . . . , tn, the permanent of A is
1
2n
n

i=1
σi

ti +
n

j=1
σj ai, j

,
where the outer sum is over all 2n
choices of (σ1, . . . , σn) ∈ {1, −1}n
.
11960. Proposed by Ulrich Abel, Technische Hochschule Mittelhessen, Friedberg,
Germany. Let m and n be natural numbers, and, for i ∈ {1, . . . m}, let ai be a real number
with 0 ≤ ai ≤ 1 . Define
f (x) =
1
x2
m

i=1
(1 + ai x)mn
− m
m

i=1
(1 + ai x)n

.
Let k be a nonnegative integer, and write f (k)
for the kth derivative of f . Show that
f (k)
(−1) ≥ 0.
http://dx.doi.org/10.4169/amer.math.monthly.124.2.179
February 2017] PROBLEMS AND SOLUTIONS 179

11961. Proposed by Mihaela Berindeanu, Bucharest, Romania. Evaluate
π/2
0
sin x
1 +
√
sin(2x)
dx.
11962. Proposed by Elton Hsu, Northwestern University, Evanston, IL. Let {Xn}n≥1 be
a sequence of independent and identically distributed random variables each taking the
values ±1 with probability 1/2. Find the distribution of the random variable

1
2
+
X1
2

1
2
+
X2
2

1
2
+ · · · .
11963. Proposed by Gheorghe Alexe and George-Florin Serban, Braila, Romania. Let
a1, . . . , an be positive real numbers with
n
k=1 ak = 1. Show that
n

i=1
(ai + ai+1)4
a2
i − ai ai+1 + a2
i+1
≥ 12n,
where an+1 = a1.
SOLUTIONS
An Inequality with Squared Tangents
11778 [2014, 456]. Proposed by Li Zhou, Polk State College, Winter Haven, FL.
Let x, y, z be positive real numbers such that x + y + z = π/2. Let f (x, y, z) =
1/(tan2
x + 4 tan2
y + 9 tan2
z). Prove that
f (x, y, z) + f (y, z, x) + f (z, x, y) ≤
9
14

tan2
x + tan2
y + tan2
z

.
Solution by Vazgen Mikayelyan, Department of Mathematics and Mechanics, Yerevan
State University, Yerevan, Armenia. Letting a = tan x, b = tan y, and c = tan z, we have
a, b, c 0, since 0 x, y, z π/2, and ab + bc + ca = 1 since
a = tan x = cot(y + z) =
1 − tan y tan z
tan y + tan z
=
1 − bc
b + c
.
By the Cauchy–Schwarz inequality,
3(a2
+ 4b2
+ 9c2
) =
3a2
b2
b2
+
11a2
b2
a2
+
b2
c2
c2
+
13b2
c2
b2
+
14c2
a2
a2
=
(3ab)2
3b2
+
(11ab)2
11a2
+
(bc)2
c2
+
(13bc)2
13b2
+
(14ca)2
14a2
≥
(14ab + 14bc + 14ca)2
3b2 + 11a2 + c2 + 13b2 + 14a2
=
142
25a2 + 16b2 + c2
.
Hence,
f (x, y, z) =
1
a2 + 4b2 + 9c2
≤
3(25a2
+ 16b2
+ c2
)
142
.
180 c
THE MATHEMATICAL ASSOCIATION OF AMERICA [Monthly 124

Adding this and the analogous inequalities obtained by cycling the variables, we obtain
f (x, y, z) + f (y, z, x) + f (z, x, y) ≤
3(42a2
+ 42b2
+ 42c2
)
142
=
9(a2
+ b2
+ c2
)
14
,
which is the desired inequality.
Editorial comment. Paolo Perfetti proved the stronger result
f (x, y, z) + f (y, z, x) + f (z, x, y) ≤
9
14
≤
9
14
(tan2
x + tan2
y + tan2
z).
Also solved by A. Ali (India), S. Baek (Korea), R. Bagby, P. P. Dályay (Hungary), O. Geupel (Germany),
P. Perfetti (Italy), R. Stong, R. Tauraso (Italy), and the proposer.
Concyclic or Collinear
11779 [2014, 456]. Proposed by Michel Bataille, Rouen, France.
Let M, A, B, C, and D be distinct points
(in any order) on a circle with center O.
Let the medians through M of triangles
MAB and MC D cross lines AB and CD at
P and Q, respectively, and meet again
at E and F, respectively. Let K be the
intersection of AF with DE, and let L be
the intersection of BF with CE. Let U
and V be the orthogonal projections of
C onto MA and D onto MB, respectively,
and assume U = A and V = B. Prove
that A, B, U, and V are concyclic if and
only if O, K, and L are collinear.
B
V
M
C
F
O
U
A
K
L
D
E
P
Q
Solution by Richard Stong, Center for Communications Research, San Diego, CA. The
problem is not quite correct. We must also assume that E and F do not coincide, hence
K and L do not coincide. (If K and L coincide, then O, K, L are clearly collinear, but
A,U, B, V need not be concyclic.)
Let R be the radius of , let N the point where lines AC and BD intersect, and let X and
Y be the reflections of O across lines AC and BD, respectively. The claim is the equivalence
of (1) A, B,U, V are concyclic, and (2) O, K, L are collinear. We show that each of these
is equivalent to (3) M is equidistant from X and Y.
(1) ⇐⇒ (3). Note that A, B,U, V are concyclic if and only if the powers from M are
equal: |MA| · |MU| = |MB| · |MV|. From trigonometry and the extended law of sines,
|MB| · |MV| = 4R2
sin
1
2
∠MOB

sin
1
2
∠MOD

cos
1
2
∠BOD

.
From the law of cosines applied to
MOY, we find that |MY|2
equals
R2
+ 4R2
cos2
1
2
∠BOD

− 4R2
cos
1
2
∠BOD

cos
1
2 (∠MOB + ∠MOD)

= R2
+ 8R2
sin
1
2
∠MOB

sin
1
2
∠MOD

cos
1
2
∠BOD

= R2
+ 2|MB| · |MV|,

and analogously, |MX|2
= R2
+ 2|MA| · |MU|. Thus, |MX| = |MY| if and only if |MA| ·
|MU| = |MB| · |MV|.
(2) ⇐⇒ (3). Lay down complex coordinates with equal to the unit circle. If a point
Z with coordinate z is on the unit circle and point W with coordinate w is any other point
in the complex plane, then the second intersection of line WZ with the unit circle has
coordinate (z − w)/(wz − 1). Hence, letting lower case letters denote coordinates of the
points with the corresponding upper case letter, we compute
e =
ab(a + b − 2m)
2ab − am − bm
and f =
cd(c + d − 2m)
2cd − cm − dm
.
By Pascal’s theorem applied to the hexagon CEDBFA, we see that K, L, and N all lie on
the line
(ce + db + f a − ed − bf − ac)z + (abd f + acef + bcde − abcf − acde − bdef )z
= ce(b + f ) + db(a + c) + f a(e + d) − ed( f + a) − bf (c + e) − ac(b + d).
If K = L, then this is the unique line through K and L. Hence, O, K, and L are collinear
if and only if
ce(b + f ) + db(a + c) + f a(e + d) − ed( f + a) − bf (c + e) − ac(b + d) = 0.
Plugging in the formulas for e and f above and factoring out (a − b)(c − d), this becomes
m

1
a
+
1
c
−
1
b
−
1
d

+ (a + c − b − d)m =
(ab − cd)(ad − bc)
abcd
.
Since x = a + c and y = b + d, this is the equation of the line perpendicular to XY through
the midpoint (a + b + c + d)/2 of XY. Hence, O, K, and L are collinear if and only if
|MX| = |MY|.
Also solved by R. Chapman (U. K.), J.-P. Grivaux (France), C. R. Pranesachar (India), and the proposer.
Altitudes of a Tetrahedron
11783 [2014, 549] and 11797 [2014, 738]. Proposed by Zhang Yun, Xi’an City, Shaanxi,
China. Given a tetrahedron, let r denote the radius of its inscribed sphere. For 1 ≤ k ≤ 4,
let hk denote the distance from the kth vertex to the plane of the opposite face. Prove that
4

k=1
hk − r
hk + r
≥
12
5
.
Solution by Roberto Tauraso, Dipartimento di Matematica, Università di Roma “Tor Ver-
gata,” Rome, Italy. The volume of the tetrahedron is given by
hk Ak
3
=
r S
3
,
where Ak is the area of the face opposite the kth vertex and S =
4
k=1 Ak is the surface
area of the tetrahedron. Hence, hk = r/tk, where tk = Ak/S. Since 0 tk 1 and the
function f (t) = (1 − t)/(1 + t) is convex on [0, +∞), we have
4

k=1
hk − r
hk + r
=
4

k=1
f (tk) ≥ 4 f

1
4
4

k=1
tk

= 4 f

1
4

=
12
5
.
182 c

Editorial comment. The problem was inadvertently repeated as Problems 11783 and 11797.
Several solvers noted that equality holds if and only if the face areas are equal. This
does not require, however, that the tetrahedron be regular. Some solvers noted that the
n-dimensional analogue of the inequality holds with lower bound n(n + 1)/(n + 2).
Also solved by A. Ali (India), S. Baek (Korea), R. Bagby, M. Bataille (France), D. M. Bătinetu-Giurgiu
T. Zvonaru (Romania), I. Borosh, R. Boukharfane (Morocco), R. Chapman (U. K.), N. Curwen (U. K.),
P. P. Dályay (Hungary), M. Dincă (Romania), D. Fleischman, H. S. Geun (Korea), O. Geupel (Germany),
M. Goldenberg M. Kaplan, J. G. Heuver (Canada), S. Hitotumatu (Japan), E. J. Ionaşcu, Y. J. Ionin,
B. Karaivanov (U.S.A.) T. S. Vassilev (Canada), O. Kouba (Syria), D. Lee (Korea), O. P. Lossers (Nether-
lands), V. Mikayelyan (Armenia), R. Nandan, Y. Oh (Korea), P. Perfetti (Italy), I. Pinelis, C. R. Pranesachar
(India), Y. Shim (Korea), J. C. Smith, R. Stong, T. Viteam (India), M. Vowe (Switzerland), T. Zvonaru
N. Stanciu (Romania), GCHQ Problem Solving Group (U. K.), Missouri State University Problem Solving
Group, University of Louisiana at Lafayette Math Club, and the proposer.
Circles around an Equilateral Triangle
11784 [2014, 549]. Proposed by Abdurrahim Yilmuz, Middle East Technical University,
Ankara, Turkey. Let ABC be an equilateral triangle with center O and circumradius r.
Given R r, let ρ be a circle about O of radius R. All points named “P” are on ρ.
(a) Prove that |PA|2
+ |PB|2
+ |PC|2
= 3(R2
+ r2
).
(b) Prove that minP∈ρ |PA| |PB| |PC| = R3
− r3
and maxP∈ρ |PA| |PB| |PC| = R3
+ r3
.
(c) Prove that the area of a triangle with sides of length |PA|, |PB|, and |PC| is
√
3
4
(R2
− r2
).
(d) Prove that if H, K, and L are the respective projections of P onto AB, AC, and BC,
then the area of triangle HKL is 3
√
3
16
(R2
− r2
).
(e) With the same notation, prove that |HK|2
+ |KL|2
+ |HL|2
= 9
4
(R2
+ r2
).
Solution by TCDmath Problem Group, Trinity College, Dublin, Ireland.
(a) We represent the points by complex numbers: A = r, B = rω, C = rω2
, O = 0,
P = z, where r 0 and ω = e2πi/3
. We compute
|PA|2
= (z − r)(z − r) = |z|2
− r(z + z) + r2
,
|PB|2
= (z − rω)(z − rω2
) = |z|2
− r(zω + zω2
) + r2
, and
|PC|2
= (z − rω2
)(z − rω) = |z|2
− r(zω2
+ zω) + r2
.
Summing these equations and using 1 + ω + ω2
= 0 yields
|PA|2
+ |PB|2
+ |PC|2
= 3(R2
+ r2
).
(b) We have
|PA| |PB| |PC| =

(z − r)(z − rω)(z − rω2
)

=

z3
− r3

.
This formula takes its maximum value R3
+ r3
when z3
= −R3
, that is, when z = Reiθ
with 3θ ≡ π (mod 2π) or when P lies on one of the altitudes of the triangle on the oppo-
site side to the vertex. It takes its minimum value R3
− r3
when z3
= R3
, that is, when P
lies on one of the three altitudes of the triangle on the same side as the vertex.
(c) Heron’s formula for the area of a triangle with sides a, b, c is
162
= 2

a2
b2
+ b2
c2
+ c2
a2

−

a4
+ b4
+ c4

=

a2
+ b2
+ c2
2
− 2

a4
+ b4
+ c4

.
In our case, we have (|PA|2
+ |PB|2
+ |PC|2
)2
= 9(R2
+ r2
)2
and

|PA|4
+ |PB|4
+ |PC|4
=

R2
+ r2
− r(z + z)
2
+

R2
+ r2
− r(zω + zω2
)
2
+

R2
+ r2
− r(zω2
+ zω)
2
= 3

R2
+ r2
2
− 2

R2
+ r2

r

z + z + zω + zω2
+ zω2
+ zω

+ r2

(z + z)2
+ (zω + zω2
)2
+ (zω2
+ zω)2

= 3

R2
+ r2
2
+ 6r2
zz = 3

R2
+ r2
2
+ 6R2
r2
.
Thus, 162
= 3(R2
+ r2
)2
− 12(R2
r2
) = 3(R2
− r2
)2
. Hence, =
√
3
4
(R2
− r2
).
(d) In coordinates (x, y), line BC has equation x = −r/2, and the projection PL has equa-
tion y = b, where b is the imaginary part of z. Thus,
L =
1
2
(−r + z − z).
To determine H, we rotate the plane with the transformation z
→ ωz and then rotate back
with z
→ ω2
z. This yields
H =
ω2
2
(−r + ωz − ω2
z) =
1
2
(−ω2
r + z − ωz).
Similarly,
K =
1
2
(−ωr + z − ω2
z).
Thus, using |1 − ω| = |1 − ω2
| = |ω − ω2
| =
√
3, we have
|H K| =
√
3
2
|z − r|, |K L| =
√
3
2
|z − ωr|, and |L H| =
√
3
2
|z − ω2
r|.
Replacing z with z in these expressions leaves the triple {|HK|, |KL|, |LH|} unchanged.
The triangle HKL is therefore similar to the triangle in part (c) with coefficient of similarity
equal to
√
3/2. It follows that the area of
HKL is 3/4 times the area of the earlier triangle,
that is, 3
√
3
16
(R2
− r2
).
(e) Similarly, comparing with the earlier triangle, |HK|2
+ |KL|2
+ |HL|2
= 9
4
(R2
+ r2
).
Editorial comment. The Editors regret that the “16” in the denominator of part (d) was
misprinted as “116.”
All parts also solved by R. Bagby, M. Bataille (France), R. Boukharfane (Morocco), R. Chapman (U. K.),
N. Curwen (U. K.), P. P. Dályay (Hungary), A. Ercan (Turkey), M. Goldenberg M. Kaplan, J.-P. Grivaux
(France), A. Habil (Syria), E. J. Ionaşcu, Y. J. Ionin, B. Karaivanov, O. Kouba (Syria), M. D. Meyerson,
J. Minkus, C. R. Pranesachar (India), J. C. Smith, R. Stong, T. Zvonaru N. Stanciu (Romania), GCHQ
Problem Solving Group (U. K.), and the proposer. Some but not all parts solved by A. Ali (India), R. B. Cam-
pos (Spain), D. Fleischman, O. Geupel (Germany), P. Nüesch (Switzerland), J. Schlosberg, C. R. Selvaraj
S. Selvaraj, T. Viteam (India), and Z. Vörös (Hungary).
A Limit of a Ratio of Logarithms
11786 [2014, 550]. Proposed by George Stoica, University of New Brunswick, Saint John,
Canada. Let x1, x2, . . . be a sequence of positive numbers such that limn→∞ xn = 0 and
limn→∞
log xn
x1+···+xn
is a negative number. Prove that limn→∞
log xn
log n
= −1.
Solution by Lixing Han, University of Michigan, Flint, MI. Suppose that
lim
n→∞
log xn/(x1 + · · · + xn) = β 0.
184 c

Then by the Stolz–Cesàro theorem, we have
lim
n→∞
log(xn+1/xn)
xn+1
= lim
n→∞
log xn+1 − log xn
xn+1
= lim
n→∞
log xn
x1 + · · · + xn
= β. (1)
This implies that xn+1 xn for large n since xn 0 and limn→∞ xn = 0 by assumption.
Since xn goes to zero as n → ∞, (1) implies limn→∞ βxn+1 = limn→∞ log(xn+1/xn) = 0.
It follows that
lim
n→∞
xn+1
xn
= 1. (2)
By the mean value theorem, there exists ζn such that
log xn+1 − log xn =
1
ζn
(xn+1 − xn),
where ζn is between xn and xn+1. Thus, for large n, xn+1/xn ζn/xn 1. From (2), we
conclude limn→∞ ζn/xn = 1. Using this result and (1) again,
lim
n→∞
log xn+1 − log xn
xn+1
= lim
n→∞
1
ζn
·
xn+1 − xn
xn+1
= lim
n→∞
xn
ζn
·
xn+1 − xn
xn xn+1
= β.
This implies that for any ∈ (0, |β|), there exists a positive integer N such that
β −
1
xn
−
1
xn+1
β +
for all n ≥ N. Summing these inequalities from n = N to N + m − 1, we obtain
(β − )m
1
xN
−
1
xN+m
(β + )m.
Dividing by N + m and taking the limit as m → ∞,
β − ≤ lim
m→∞

1
(N + m)xN
−
1
(N + m)xN+m

≤ β + .
Since limm→∞
1
(N+m)xN
= 0,
1
β +
≤ − lim
m→∞
(N + m)xN+m = − lim
n→∞
nxn ≤
1
β −
.
Let approach 0 to obtain limn→∞ nxn = −1/β 0. Thus,
lim
n→∞
log(nxn) = lim
n→∞
(log n + log xn) = log

−
1
β

.
However, if this holds, then, since log n → ∞, it must be the case that
lim
n→∞
log n + log xn
log n
= lim
n→∞
log(− 1
β
)
log n
= 0.
Hence,
lim
n→∞

1 +
log xn
log n

= 0,
so limn→∞ log xn/ log n = −1.

Also solved by P. Bracken, P. P. Dályay (Hungary), P. J. Fitzsimmons, E. J. Ionaşcu, O. Kouba (Syria),
J. H. Lindsey II, O. P. Lossers (Netherlands), M. Omarjee (France), T. Persson M. P. Sundqvist (Sweden),
I. Pinelis, J. C. Smith, A. Stenger, R. Stong, and the proposer.
Sum of Medians of a Triangle
11790 [2014, 648]. Proposed by Arkady Alt, San Jose, CA, and Konstantin Knop, St. Peters-
burg, Russia. Given a triangle with semiperimeter s, inradius r, and medians of length ma,
mb, and mc, prove that ma + mb + mc ≤ 2s − 3(2
√
3 − 3)r.
Solution by James Christopher Smith, Knoxville, TN. Write R for the circumradius. We use
two inequalities. The first is
(ma + mb + mc)2
≤ 4s2
− 16Rr + 5r2
,
due to Xiao-Guang Chu and Xue-Zhi Yang. (See J. Liu, “On an inequality for the medians
of a triangle,” Journal of Science and Arts, 19 (2012) 127–136.) The second is
s ≤ (3
√
3 − 4)r + 2R,
known as Blundon’s inequality. (See problem E1935, this MONTHLY, 73 (1966) 1122.)
Write u = 2
√
3 − 3. From Blundon’s inequality,
(2s − 3ur)2
= 4s2
− 12sur + 9u2
r2
≥ 4s2
− 12ur

(3
√
3 − 4)r + 2R

+ 9u2
r2
= 4s2
− 24uRr +

9u2
− 12u(3
√
3 − 4)

r2
= 4s2
− 16Rr + (16 − 24u)Rr + 3u(7 − 6
√
3 )r2
.
Next, we use Euler’s inequality R ≥ 2r to get

2s − 3ur
2
≥ 4s2
− 16Rr + (16 − 24u)2r2
+ 3u(7 − 6
√
3 )r2
= 4s2
− 16Rr + 5r2
,
which is greater than or equal to (ma + mb + mc)2
by the Chu–Yang inequality.
Also solved by R. Boukharfane (Canada), O. Geupel (Germany), O. Kouba (Syria), R. Tauraso (Italy), M. Vowe
(Switzerland), and T. Zvonaru N. Stanciu (Romania).
A Middle Subspace
11792 [2014, 648]. Proposed by Stephen Scheinberg, Corona del Mar, CA. Show that every
infinite-dimensional Banach space contains a closed subspace of infinite dimension and
infinite codimension.
Solution by University of Louisiana at Lafayette Math Club, Lafayette, LA. Let V be an
infinite-dimensional normed vector space (we do not require completeness). We construct a
sequence of linearly independent vectors v0, v1, . . . in V and a sequence of bounded linear
functionals λ0, λ1, . . . such that λi (vj ) = δi, j for all nonnegative integers i and j. Choose
a nonzero v0 ∈ V . By the Hahn–Banach theorem, there is a bounded linear functional λ0
on V with λ0(v0) = 1. Suppose that nonzero vectors v0, . . . , vk ∈ V and bounded linear
functionals λ0, . . . , λk have been defined such that λi (vj ) = δi, j for i, j ∈ {1, . . . , k}. The
vector subspace
k
i=1 ker λi has infinite dimension since it has finite codimension in V ,
which is infinite-dimensional. In particular, there exists nonzero vk+1 ∈
k
i=1 ker λi . The
186 c

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