Problems and Solutions
Source: The American Mathematical Monthly, Vol. 123, No. 10 (December 2016), pp. 10501057
Published by: Mathematical Association of America
Stable URL: http://www.jstor.org/stable/10.4169/amer.math.monthly.123.10.1050
Accessed: 09-01-2017 22:55 UTC
JSTOR is a not-for-profit service that helps scholars, researchers, and students discover, use, and build upon a wide range of content in a trusted
digital archive. We use information technology and tools to increase productivity and facilitate new forms of scholarship. For more information about
JSTOR, please contact [email protected].
Your use of the JSTOR archive indicates your acceptance of the Terms & Conditions of Use, available at
http://about.jstor.org/terms
Mathematical Association of America is collaborating with JSTOR to digitize, preserve and extend access
to The American Mathematical Monthly
This content downloaded from 129.59.95.115 on Mon, 09 Jan 2017 22:55:59 UTC
All use subject to http://about.jstor.org/terms
PROBLEMS AND SOLUTIONS
Edited by Gerald A. Edgar, Doug Hensley, Douglas B. West
with the collaboration of Itshak Borosh, Paul Bracken, Ezra A. Brown, Randall
Dougherty, Tamás Erdélyi, Zachary Franco, Christian Friesen, Ira M. Gessel, László
Lipták, Frederick W. Luttmann, Vania Mascioni, Frank B. Miles, Steven J. Miller,
Mohamed Omar, Richard Pfiefer, Dave Renfro, Cecil C. Rousseau, Leonard Smiley,
Kenneth Stolarsky, Richard Stong, Walter Stromquist, Daniel Ullman, Charles Vanden
Eynden, and Fuzhen Zhang.
Proposed problems should be submitted online via www.americanmathematical
monthly.submittable.com. Proposed solutions to the problems below should be submitted on or before April 30, 2017 at the same link. More detailed instructions are
available online. Solutions to problems numbered 11921 or lower should continue
to be submitted via email to [email protected]. Proposed problems
must not be under consideration concurrently to any other journal and must not 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
11943. Proposed by Keith Kearnes, University of Colorado, Boulder, CO, and Greg
Oman, University of Colorado, Colorado Springs, CO. Let X be a set, and let F be
a collection of functions f from X into X . A subset Y of X is closed under F if
f (y) ∈ Y for all y ∈ Y and f in F . With the axiom of choice given, prove or disprove:
There exists an uncountable collection F of functions mapping Z+ into Z+ such that
(a) every proper subset of Z+ that is closed under F is finite, and
(b) for every f ∈ F , there is a proper infinite subset Y of Z+ that is closed under
F \{ f }.
11944. Proposed by Yury Ionin, Central Michigan University, Mount Pleasant, MI. Let
n be a positive integer, and let [n] = {1, . . . , n}. For i ∈ [n], let Ai , Bi , Ci be disjoint
sets such that Ai ∪ Bi ∪ Ci = [n] − {i} and |Ai | = |Bi |. Suppose also that
|Ai ∩ B j | + |Bi ∩ C j | + |Ci ∩ A j | = |Bi ∩ A j | + |Ci ∩ B j | + |Ai ∩ C j |
for i, j ∈ [n]. Prove that i ∈ A j if and only if j ∈ Ai and, likewise, for the Bs and Cs.
11945. Proposed by Martin Lukarevski, University “Goce Delcev,” Stip, Macedonia.
Let a, b, and c be the lengths of the sides of triangle ABC opposite A, B, and C,
respectively, and let wa , wb , wc be the lengths of the corresponding angle bisectors.
Prove
√
b
c
a
+
+
≥ 2 3.
wa
wb
wc
11946. Proposed by Moubinool Omarjee, Lycée Henri IV, Paris, France. Let f be
a twice differentiable function from [0, 1] to R with f continuous on [0, 1] and
2/3
f (x) d x = 0. Prove
1/3
http://dx.doi.org/10.4169/amer.math.monthly.123.10.1050
1050
c THE MATHEMATICAL ASSOCIATION OF AMERICA [Monthly 123
This content downloaded from 129.59.95.115 on Mon, 09 Jan 2017 22:55:59 UTC
All use subject to http://about.jstor.org/terms
2
1
f (x) d x
4860
1
f (x) d x.
≤ 11
0
0
11947. Proposed by George Stoica, University of New Brunswick, Saint John, Canada.
Let n be a positive integer, and let z 1 , . . . , z n be the zeros in C of z n + 1. For a > 0,
prove
n
1 + a 2 + · · · + a 2(n−1)
1
1
=
.
n k=1 |z k − a|2
(1 + a n )2
11948. Proposed by Navid Safaei, Sharif University of Technology, Tehran, Iran. Find
all surjective functions f : R → R+ such that (1) f (x) ≤ x + 1 for f (x) ≥ 1, (2)
f (x) = 1 for x = 0, and (3) for x, y ∈ R,
f (x f (y) + y f (x) − x y) = f (x) f (y).
11949. Proposed by Eugen J. Ionascu, Columbus State University, Columbus, GA.
Show that there exists a unique function f from R to R such that f is differentiable,
2 cos(x + f (x)) − cos x = 1 for all real x, and f (π/2) = −π/6.
SOLUTIONS
Flett’s Mean Value Theorem
11814 [2015, 76]. Proposed by Cezar Lupu, University of Pittsburgh, Pittsburgh, PA.
Let φ be a continuously differentiable function from [0, 1] into R, with φ(0) = 0 and
φ(1) = 1, and suppose that φ (x) = 0 for 0 ≤ x ≤ 1. Let f be a continuous function
1
1
from [0, 1] into R such that 0 f (x) d x = 0 φ(x) f (x) d x. Show that there exists t
t
with 0 < t < 1 such that 0 φ(x) f (x) d x = 0.
Solution by New York Math Circle, NY. Define
s
φ(x) f (x) d x − φ(s)
h(s) =
0
s
f (x) d x.
0
Note that h(0) = 0 = h(1). From Rolle’s theorem, we obtain h (c) = 0 for some c ∈
(0, 1). Also, we compute
s
f (x) d x
h (s) = −φ (s)
0
and, in particular, h (0) = 0. Since φ (s) = 0 for all s ∈ (0, 1), the inverse function
φ −1 (s) exists and is differentiable on (0, 1). Letting H (s) = h(φ −1 (s)), we see that
H (s) =
h (φ −1 (s))
.
φ (φ −1 (s))
Applying Flett’s mean value theorem [Math. Gaz. 42 (1958) 38–39] to the function H
on the interval [0, φ(c)], we have
H (T ) − H (0)
= H (T )
T −0
December 2016]
PROBLEMS AND SOLUTIONS
This content downloaded from 129.59.95.115 on Mon, 09 Jan 2017 22:55:59 UTC
All use subject to http://about.jstor.org/terms
1051
for some T ∈ (0, φ(c)) ⊂ (0, 1). Setting t = φ −1 (T ) ∈ (0, 1), this becomes
t
t
t
h (t)
1
h(t)
= =−
φ(x) f (x) d x −
f (x) d x =
f (x) d x.
φ(t) 0
φ(t)
φ (t)
0
0
t
Thus 0 φ(x) f (x) d x = 0, as desired.
Also solved by K. F. Andersen (Canada), R. Bagby, M. W. Botsko, B. S. Burdick, R. Chapman (U. K.),
P. P. Dályay (Hungary), M. Dan-Ştefan & F. Cătălin-Emil & O. Alexandru (Romania), L. Giugiuc (Romania),
D. Hancock, E. A. Herman, O. Kouba (Syria), J. K. Lindsey II, O. P. Lossers (Netherlands), A. Mingarelli &
J. M. Pacheco & Á. Plaza (Spain), P. Perfetti (Italy), I. Pinelis, D. Ritter, B. Schmuland (Canada), R. Stong, R.
Tauraso (Italy), T. P. Turiel, T. Wiandt, M. Wildon (U. K.), and the proposer.
Pascal’s Theorem
11816 [2015, 76]. Proposed by Sabin Tabirca, University College Cork, Cork, Ireland.
Let ABC be an acute triangle, and let B1 and C1 be the points where the altitudes from
B and C intersect the circumcircle. Let X be a point on arc BC, and let B2 and C2
denote the intersections of XB1 with AC and XC1 with AB, respectively. Prove that the
line B2 C2 contains the orthocenter of ABC.
Solution by Adnan Ali, A.E.C.S.-4, Mumbai, India. The claim holds not only for the circumcircle of ABC but also for any circumconic of the triangle—i.e. a conic circumscribing the triangle—as this problem is a special case of Pascal’s theorem, according
to which, if ABCDEF is a hexagon with vertices on a conic, then the intersections of
lines AB with ED, AF with CD, and EF with CB are collinear. (A point at infinity is
allowed.)
Also solved by M. Atasever (Turkey), M. Bataille (France), B. S. Burdick, J. Cade, R. B. Campos (Spain), R.
Chapman (U. K.), P. P. Dályay (Hungary), M. Dan-Ştefan & O. Alexandru & F. Cătălin-Emil (Romania), P.
De (India), O. Faynshteyn, D. Fleischman, O. Geupel (Germany), M. Goldenberg & M. Kaplan, J.-P. Grivaux
(France), J. G. Heuver (Canada), S. Hong (Korea), E. J. Ionaşcu, Y. J. Ionin, I. M. Isaacs, O. Kouba (Syria),
G. Lord, O. P. Lossers (Netherlands), J. Minkus, M. A. Shayib, N. Stanciu & T. Zvonaru (Romania), R. Stong,
T. Viteam (India), Z. Vörös (Hungary), T. Wiandt, GCHQ Problem Solving Group (U. K.), Missouri State
University Problem Solving Group, and the proposer.
Cycle Covers for Infinite Complete Graphs
11817 [2015, 175]. Proposed by Mohammed Jahaveri, Siena College, Loudonville, NY.
A cycle double cover of a graph is a collection of cycles that, counting multiplicity,
includes every edge exactly twice. Let X be an infinite set and let K X be the complete
graph on X . Construct a cycle double cover for X .
Solution I by I. M. Isaacs, Berkeley, CA. We construct a set of triangles covering
each edge exactly once. Taking each triangle twice produces a cycle double cover.
We may replace X by another set of the same cardinality, so we consider the set S of
all nonempty finite subsets of X . For each edge AB in K S , let C = AB (the symmetric difference), and use the triangle with vertices A, B, C. Since BC = A and
CA = B, each edge lies in exactly one such triangle.
Solution II by Jerrold Grossman and László Lipták, Oakland University, Rochester,
MI. We partition the edges into triangles. Use the axiom of choice to well order the
edges of K X using an ordinal that is also a cardinal. Transfinitely perform the following operation as long as there remains an edge not yet covered: For the least such
uncovered edge uv, choose a vertex w not yet in any triangle, and add uvw to the set
of triangles. Such a vertex w exists because the cardinality of the set of vertices used
so far in the process is less than the cardinality of X .
1052
c THE MATHEMATICAL ASSOCIATION OF AMERICA [Monthly 123
This content downloaded from 129.59.95.115 on Mon, 09 Jan 2017 22:55:59 UTC
All use subject to http://about.jstor.org/terms
Editorial comment. The method of Grossman and Lipták can be used to partition edges
of K X into copies of G for any finite graph G.
Also solved by E. Bojaxhiu (Albania) & E. Hysnelaj (Australia), B. Burdick, R. Chapman (U. K.), B. Karaivanov
& T. Vassilev (Canada), R. Stong, FAU Problem Solving Group, and the proposer.
Some Trig for the Nagel Cevians
11818 [2015, 175]. Proposed by Oleh Faynshteyn, Leipzig, Germany. Let ABC be a triangle and let A1 , B1 , and C1 be the points on sides opposite A, B, and C respectively
at which the ecircles of the triangle are tangent to those sides. Let R and r be the circumradius and inradius of the triangle. Let the name of a vertex of ABC or of A1 B1 C1
also stand for the radian measure of the corresponding angle. Prove that, wherever the
expression is defined,
6R
cot A1 + cot(A/2) cot B1 + cot(B/2) cot C1 + cot(C/2)
+
+
=
.
cot A
cot B
cot C
r
Solution by P. Nüesch, Switzerland. In fact, more is true: Each term on the left side
of the identity equals 2R/r . Write a, b, c for the side lengths of ABC, s for the
semiperimeter, and F for the area. Write u, v, w for the side lengths of A1 B1 C1 and
F1 for the area. Now
2R
F
.
=
F1
r
(1)
1
4s(s − a)
cot(A/2)
.
=
+1 = 2
cot A
cos A
b + c2 − a 2
(2)
By the law of cosines,
The modified cosine laws
cot A =
b2 + c2 − a 2
,
4F
and
cot A1 =
v 2 + w2 − u 2
4F1
together with (1) give us
2R v 2 + w2 − u 2
cot A1
=
.
cot A
r b2 + c2 − a 2
(3)
Now we have to prove (3) + (2) = 2R/r , or equivalently,
2R 2
(b − v 2 ) + (c2 − w2 ) − (a 2 − u 2 ) = 4s(s − a).
r
Observe that b2 − v 2 = 2(s − c)(s − a)(1 + cos B) = 4(s − c)(s − a) cos2 (B/2) =
bF/R and similarly for the other two sides. Therefore, as required,
2R bF
cF
aF
2R
+
−
=
[b + c − a] = 2s · 2(s − a) = 4s(s − a).
r
R
R
R
r
Editorial comment. Lines AA1 , BB1 , and CC1 are the Nagel cevians of the triangle.
Also solved by A. Alt, R. Bagby, R. Chapman (U. K.), H. Y. Far, M. E. Kuczma (Poland), J. C. Smith, R.
Stong, H. Widmer, and the proposer.
December 2016]
PROBLEMS AND SOLUTIONS
This content downloaded from 129.59.95.115 on Mon, 09 Jan 2017 22:55:59 UTC
All use subject to http://about.jstor.org/terms
1053
Twin Hölders
11819 [2015, 175]. Proposed by Cezar Lupu, University of Pittsburgh, Pittsburgh, PA.
Let f be a continuous, nonnegative function on [0, 1]. Show that
1
1
1
3
2
2
f (x) d x ≥ 4
x f (x) d x
x f (x) d x .
0
0
0
Solution by Radouan Boukharfane, Université du Poitiers, Chasseneuil, France. We
apply Hölder’s inequality twice
1
2/3 1
1/3
1
x 2 f (x) d x ≤
x3 dx
f 3 (x) d x
0
1
0
x f 2 (x) d x ≤
0
Now multiply the inequalities
1
1
2
2
x f (x) d x x f (x) d x ≤
1
0
0
0
2/3
1
f 3 (x) d x
x3 dx
0
0
1/3 1
.
0
3
1
1
f (x) d x =
4
1
3
x dx
0
f (x) d x .
3
0
Editorial comment. Several solvers proved generalizations. For example,
argument
the
a
b
(x)g
(x) d x
above,
using
the
conjugate
exponents
(a
+
b)/a
and
(a
+
b)/b,
yields
f
a+b
a+b
b
a
f (x)g (x) d x ≤ f (x) d x g (x) d x.
Also solved by R. A. Agnew, A. Alt, T. Amdeberhan & V. H. Moll, K. F. Andersen (Canada), R. Bagby,
M. Bataille (France), P. Bracken, M. A. Carlton, R. Chapman (U. K.), H. Chen, L. V. P. Cuong (Vietnam),
P. J. Fitzsimmons, W. R. Green, N. Grivaux (France), E. A. Herman, B. Karaivanov (USA) & T. S. Vazzilev
(Canada), O. Kouba (Syria), M. E. Kuczma (Poland), K.-W. Lau (China), J. H. Lindsey II, P. W. Lindstrom,
M. Omarjee (France), X. Oudot (France), P. Perfetti (Italy), Á. Plaza & F. Perdomo (Spain), K. Schilling, J. G.
Simmonds, J. C. Smith, A. Stenger, R. Stong, R. Tauraso (Italy), J. Vinuesa (Spain), H. Wang & J. Wojdylo,
G. White, Q. Zhang (China), Z. Zhang (China), NSA Problems Group, and the proposer.
Noetherian Subrings
11820 [2015, 175]. Proposed by Alborz Azarang, Shahid Chamran University of
Ahvaz, Ahvaz, Iran. Let K be a field and let R be a subring of K [X ] that contains K .
Prove that R is noetherian, that is, that every ascending chain of ideals in R terminates.
Solution by the National Security Agency Problems Group, Fort Meade, MD. Since a
finitely generated K -algebra is a quotient of K [x1 , . . . , xn ] for some n and, hence, is
noetherian, it suffices to show that R is finitely generated as a K -algebra. We use a
lemma of independent interest.
Lemma. Any set S of nonnegative integers that is closed under addition is finitely
generated: Thatis, there are elements d1 , . . . , dn ∈ S such that every s ∈ S can be
written as s = nk=1 ek dk for some nonnegative integers e1 , . . . , en .
Proof. This is clear if S is empty or equals {0}. Otherwise, let n be the least positive
integer in S. For 1 ≤ i < n, let di be the least element of S congruent to i (modulo
n), or di = 0 if S has no such element. We claim that {d1 , . . . , dn−1 , n} generates S.
If s ∈ S, then s ≡ di mod n for some i. Also, s ≥ di . Hence, s = di + kn for some
nonnegative integer k.
Now let R be a subring of K [X ] that contains K . Let S be the degrees of the
elements of R; note that S is closed under addition. By the lemma, there are integers d1 , . . . , dn that generate S. For a ≤ i ≤ n, let f i be a monic polynomial in
1054
c THE MATHEMATICAL ASSOCIATION OF AMERICA [Monthly 123
This content downloaded from 129.59.95.115 on Mon, 09 Jan 2017 22:55:59 UTC
All use subject to http://about.jstor.org/terms
R of degree di . Using induction on the degree m of a polynomial in R, we prove
R = K [ f 1 , . . . , f n ]. For m = 0, note that the constant polynomials are in R. For
m > 1, let a be the leading
coefficient of a polynomial f in R. There are integers
e
e
ei ≥ 0 such that m = nk=1 ek dk . Let g = f − a f 1 1 f 2 2 · · · f nen . Note that g is in R and
has degree less than m. By the induction hypothesis, g ∈ K [ f 1 , . . . , f n ]. Hence, also,
f ∈ K [ f 1 , . . . , f n ], as desired.
Editorial comment. Various solvers used theorems from commutative algebra such as
the Eakin–Nagata theorem, the Artin–Tate theorem, and the Hilbert basis theorem, as
well as the chicken McNugget theorem, which is also known as the Frobenius coin
problem from number theory.
Also solved by A. J. Bevelacqua, T. Borislav (Canada) & V. Karaivanov, N. Caro (Brazil), R. Chapman (U. K.),
I. M. Isaacs, J. H. Lindsey II, F. Perdomo & A. Francisco (Spain), J. C. Smith, R. Stong, D. Ware, and the
proposer.
When a Composition of Polynomials Is Real
11822 [2015, 176]. Proposed by George Stoica, University of New Brunswick, Saint
John, Canada. Call a polynomial real if all its coefficients are real. Let P and Q be
polynomials with complex coefficients such that the composition P ◦ Q is real. Show
that if the leading coefficient of Q and its constant term are both real then P and Q are
real.
Solution by Borislav Karaivanov, Sigma Space, Lanham, MD & Tzvetalin Vassilev,
Nipissing University, North
q Bay, iON, Canada. Let P and Q be defined as P(x) =
p
i
a
x
and
Q(x)
=
i=0 i
i=0 bi x with bq and b0 real. Since both bq and the coefficient a p bqp of x pq in P(Q(x)) are real, it can be concluded that a p is real.
We first claim that Q is real. If not, then let k be the largest index for which bk is
not real. The coefficient of x q( p−1)+k in P(Q(x)) is real and has the form a p bqp−1 bk +
M, where M is a polynomial expression in a p and bi , for k < i ≤ q, with integer
coefficients, and thus real. Hence, bk must be real. This contradiction shows that Q is
real.
Next, we claim that P is real. If not, then let k be the largest index for which ak is not
real. Consider the coefficient of x qk in P(Q(x)). By the premise of the problem, it is
real. On the other hand, it has the form ak bqk + N , where N is a polynomial expression
in the real variables ak+1 , . . . , a p and b0 , . . . , bq with integer coefficients . Therefore,
there is no such ak , and P is real.
Also solved by B. Bekker (Russia) & Y. J. Ionin (USA), A. J. Bevelacqua, N. Caro (Brazil), R. Chapman
(U.K.), P. De, B. Sury (India) & N. V. Tejaswi (Netherlands), D. Fleischman, J.-P. Grivaux (France), E. A.
Herman, P. W. Lindstrom, R. Stong, R. Tauraso (Italy), N. Thornber, J. Van Hamme (Belgium), T. Viteam
(Japan), and the proposer.
Inversion in a Circle?
11823 [2015, 176]. Proposed by Sabin Tabirca, University College Cork, Cork,
Ireland. Let P be a point inside a circle C.
(a) Prove that there exists a point P outside C such that, for all chords XY of C
through P, (|XP | + |YP |)/|XY| is the same. (Here, |UV| denotes the distance from U
to V .)
(b) Is P unique?
Solution by Ahmad Habil, Damascus University, Damascus, Syria. Let O denote the
center of C, r the radius of C, and p the distance of P from O. We must exclude the
December 2016]
PROBLEMS AND SOLUTIONS
This content downloaded from 129.59.95.115 on Mon, 09 Jan 2017 22:55:59 UTC
All use subject to http://about.jstor.org/terms
1055
case P = O (i.e., p = 0) because in that case there is no suitable P at all. We (1)
show this then (2) show that, for any other P inside C, the image of P under inversion
in C can serve as P and then finally (3) show that no other point can play the role of
P (so in answer to (b), P is unique).
(1) Suppose P = O, and let Q be any “trial” point outside C. We will demonstrate that Q cannot meet the requirement for P by finding two chords X 1 Y2 and X 2 Y2
through P for which the ratios (|X j Q| + |Y j Q|)/|X j Y j | are unequal. Let q = |OQ|.
diamLet X 1 , Y1 be the diameter lying along O Q and let X 2 , Y2 be the perpendicular
eter. Now (|X 1 Q| + |Y1 Q|)/|X 1 Y1 | = qr , but (|X 2 Q| + |Y2 Q|)/|X 2 Y2 | = r 2 + q 2 /r
> qr .
(2) Consider P = O, so p = 0. Let P be the image of P under inversion in C.
This is the point on ray OP such that pp = r 2 , where p = |OP |. When T is any
point on C (but not on line OP), triangles TOP and P O T are similar, since they have
a common angle at O and |TP|/|OP| = |OP |/|OT| or rp = pr from the definition of
p . Thus, |TP |/|OT| = |TP|/|OP|, or |TP | = rp |TP|.
Now let XY denote any chord through P. We have both |XP | + rp |XP| and |YP | =
r
|YP|. Adding, |XP | + |YP | = rp (|XP| + |YP|). Since XPY is a straight line, |XP| +
p
|YP| = |XY|. We conclude (|XP | + |YP |)/|XY| = rp .
Note that the constant value of (|XP | + |YP |)/|XY| must be pr , which equals rp and
is the cosecant of half the angle intercepted by C at P .
(3) Now let Q be a point outside C such that (|XQ | + |YQ |)/|XY| is constant for
all chords XY containing P. We must show that Q is the inversion image of P in C,
that is, the point P from part (2). Let Q be the image of Q under inversion in C.
Thus, qq = r 2 , where q = |QO| and q = |Q O|. Inversion is self-dual, so our claim
is that Q = P.
Applying part (2) starting with Q, we obtain that (|XQ | + |YQ |)/|XY| is constant
for all chords through Q. Consider the diameter X 1 Y1 containing Q and the chord
X 2 Y2 lying along line PQ . Label them so that X i is closer than Yi to Q in each case.
(These chords may be the same; in fact, we prove that they are.)
Now consider a chord X Y through both P and Q. (If P = Q, as we will show,
then there are infinitely many such chords, but in any case, there is at least one.)
Because both X Y and X 1 Y1 include Q, we have (|X Q | + |Y Q |)/|X Y |
= (|X 1 Q | + |Y1 Q |)/|X 1 Y1 |. Next, since both X Y and X 2 Y2 include P, we have
(|X Q | + |Y Q |)/|X Y | = (|X 2 Q | + |Y2 Q |)/|X 2 Y2 |. Therefore,
|X 1 Q | + |Y1 Q |
|X 2 Q | + |Y2 Q |
=
.
|X 1 Y1 |
|X 2 Y2 |
Using |Y1 Q | + |X 1 Q | + |X 1 Y1 | and |Y2 Q | = |X 2 Q | + |X 2 Y2 |, we get (2|X 1 Q |
+ |X 1 Y1 |)/|X 1 Y1 | = (2|X 2 Q | + |X 2 Y2 |)/|X 2 Y2 |. Subtracting 1 and dividing by 2
yields |X 1 Q |/|X 1 Y1 | = |X 2 Q |/|X 2 Y2 |. Inverting these ratios and adding 1 gives
(|X 1 Q | + |X 1 Y1 |)/|X 1 Q | = (|X 2 Q | + |X 2 Y2 |)/|X 2 Q |. Since Q X 1 Y1 and Q X 2 Y2
are straight lines, |Y1 Q |/|X 1 Q | = |Y2 Q |/|X 2 Q |, or equivalently, |Y1 Q |/|Y2 Q |
= |X 1 Q |/|X 2 Q |.
By the concurrent chords theorem, |Y2 Q |/|Y1 Q | = |X 1 Q |/ X 2 Q |. Hence, |Y1 Q |/
|Y2 Q | = |Y2 Q |/|Y1 Q |, so |Y2 Q | = |Y1 Q |, and in turn |X 2 Q | = |X 1 Q |. This
implies that X 1 Y1 and X 2 Y2 are the same chord and, therefore, that P and Q are the
same point. Finally, Q = P , and so the image of P under inversion is the unique
point with the desired constant ratio property.
1056
c THE MATHEMATICAL ASSOCIATION OF AMERICA [Monthly 123
This content downloaded from 129.59.95.115 on Mon, 09 Jan 2017 22:55:59 UTC
All use subject to http://about.jstor.org/terms
Editorial comment. Several solvers noted that circle C is the Circle of Apollonius
determined by P and its image under inversion in C, using the ratio (r − p)/(r + p)
to get started. This observation provides another way to prove (b), that the image of P
under inversion in C is a suitable P .
Also solved by R. Bagby, M. Bataille (France), E. Bojaxhiu (Albania)& E. Hysnelaj (Australia), J. Cade, R.
Chapman (U. K.), W. J. Cowieson, E. A. Herman, L. R. King, M. E. Kuczma (Poland), G. Lord, J. Schlosberg,
J. C. Smith, N. Stanciu & T. Zvonaru (Romania), R. Stong, E. A. Weinstein, and the proposer.
A Binomial Coefficient Inequality
11826 [2015, 284]. Proposed by Michel Bataille, Rouen, France. Let m and n be
positive integers with m ≤ n. Prove that
n
n m+k−1 2 m+n 2
4n+1−k
≥
.
m−1
k
k=m
k=m
Solution by Timothy Woodcock,
College, Easton, MA. Equality holds
2m−1Stonehill
when n = m since 2m
=
2
.
Now
suppose
n > m, and inductively assume
m
m−1
n−1 n−k m+k−12 n−1 m+n−12
≥ k=m
. We have
k=m 4
m−1
k
n
k=m
n+1−k
4
m+k−1
m−1
2
2
n−1
m+n−1 2
n−k m + k − 1
=4
+4
4
m−1
m−1
k=m
n−1 n−1 m+n−1 2
m+n−1 2
m+n−1 2
≥4
+4
=4
.
m−1
k
k
k=m
k=m−1
m+n−12
2
≥ nk=m m+n
. Since (x + y)2 ≤
It now suffices to prove 4 n−1
k=m−1
k
k
2(x 2 + y 2 ) for x, y ∈ R,
n n m+n 2 m+n−1
m+n−1 2
=
+
k
k−1
k
k=m
k=m
n
m+n−1 2
m+n−1 2
≤
2
+
k−1
k
k=m
n−1 n m+n−1 2
m+n−1 2
m+n−1 2
=4
+4
=4
.
k
k
n
k=m
k=m
Editorial comment. Allen Stenger notes that (x + y) p ≤ 2 p−1 (x p + y p ), valid for
x, y > 0 and p > 1, may be used in place of (x + y)2 ≤ 2(x 2 + y 2 ) to yield the generalization
p n
n m+n p
p(n+1−k) m + k − 1
2
≥
.
m−1
k
k=m
k=m
Also solved by R. Chapman (U. K.), J. H. Lindsey II, J. C. Smith, A. Stenger, R. Stong, R. Tauraso (Italy), and
the proposer.
December 2016]
PROBLEMS AND SOLUTIONS
This content downloaded from 129.59.95.115 on Mon, 09 Jan 2017 22:55:59 UTC
All use subject to http://about.jstor.org/terms
1057