THE AMERICAN MATHEMATICAL MONTHLY

Tạp chí toán học AMM của Mỹ

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 theMONTHLY

problems address on the inside front cover Submitted solutions should arrive at that address before September 30, 2008 Additional information, such as gen-eralizations and references, is welcome The problem number and the solver’s name and address should appear on each solution An asterisk (*) after the num-ber of a problem or a part of a problem indicates that no solution is currently available.

PROBLEMS

11362 Proposed by David Callan, University of Wisconsin, Madison, WI A bit string

arc diagram is an undirected graph in which the vertices are the positions in a single

string of bits and the edges are called arcs due to the visual representation in which they are drawn joining positions in the string To be a good diagram, arcs must occur

only between unequal bits, and each bit may be the left endpoint of at most one arc Thus, the first diagram is good but, for two reasons, the second is not

1 1 0 0 1 1 0 1 1 0 0 1 1 0

There are six good diagrams on two bits, four with no arc and two with a single arc

How many good diagrams are there on n bits?

11363 Proposed by Oleh Faynshteyn, Leipzig, Germany Let m a , m b , and m c be the

lengths of the medians of a triangle T Similarly, let I a , I b , I c , h a , h b , and h c be the

lengths of the angle bisectors and altitudes of T , and let R, r , and S be the circumra-dius, inracircumra-dius, and area of T Show that

I a I b

I c

+ I b I c

I a

+ I c I a

I b

≥ 3(2R − r),

and

m a I b

h c

+m b I c

h a

+ m c I a

h b

≥ 35/4√

S

11364 Proposed by P´al P´eter D´alyay, Szeged, Hungary Let p be a prime greater than

3, and t the integer nearest p /6.

(a) Show that if p = 6t + 1, then

(p − 1)! 2t−1

j=0

(−1) j

 1

3 j+ 1+

1

3 j+ 2



≡ 0 (mod p).

(b) Show that if p = 6t − 1, then

(p − 1)!



2t−1



j=0

(−1) j

3 j+ 1+

2t−2



j=0

(−1) j

3 j+ 2



≡ 0 (mod p).

11365 Proposed by Aviezri S Fraenkel, Weizmann Institute of Science, Rehovot,

Israel Let t be a positive integer Let γ =√t2+ 4, α = 1

2(2 + γ − t), and β =

1

2(2 + γ + t) Show that for all positive integers n,

nβ = (nα + n(t − 1))α + 1 = (nα + n(t − 1) + 1)α − 1.

11366 Proposed by Nicolae Anghel, University of North Texas, Denton, TX Let φ :

R → R be a continuously differentiable function such that φ(0) = 0 and φis strictly

increasing For a > 0, let C a denote the space of all continuous functions from[0, a]

intoR, and for f ∈ C a , let I ( f ) =a

x=0(φ(x) f (x) − xφ( f (x))) dx Show that I has

a finite supremum on C a and that there exists an f ∈ C a at which that supremum is attained

11367 Proposed by Andrew Cusumano, Great Neck, NY Let x1 =√1+ 2, x2 =



1+ 2√1+ 3, and in general, let x n+1 be the number obtained by replacing the

innermost expression (1 + (n + 1)) in the nested square root formula for x n with

1+ (n + 1)√1+ (n + 2) Show that

lim

n→∞

x n − x n−1

x n+1− x n

= 2.

11368 Proposed by Wei-Dong Jiang, Weihai Vocational College, Weihai, ShanDong,

China For a triangle of area 1, let a, b, and c be the lengths of its sides Let s =

(a + b + c)/2 Show that the weighted average of (s − a)2, (s − b)2, and (s − c)2,

weighted by the radian measure of the angles opposite a, b, and c respectively, is at

leastπ/√3

SOLUTIONS

Generating Functions and Hypergeometric Series and Continued Fractions!

11198 [2006, 79] Proposed by P R Parthasarathy, India Institute of Technology

Madras, Chennai, India Let f (k) = 1 + k if k is odd, f (k) = 1 + k/2 if k is even.

Show that

0



i1=0

f (i1)1+i1

i2=0

f (i2)1+i2

i3=0

f (i3) · · ·

1+in−1

i n=0

f (i n ) =

n



m=1

m



k=1

(−1) m −k m

k k n

Composite solution by the proposer and the editors We begin by forming a generating

function from each side Multiply by (−1) n q n , sum over n ≥ 1, and use 1 for the constant term It suffices to show that the two resulting generating functions are the same We start with the one on the left

We first convert the power series to a continued fraction With a= (a0, a1, ), let

p n (a) be the coefficient of q nin the expression

1

1+ a0q

1+ a1q

1+

= 1

1+

a0q

1+

a1q

1+ · · · =

∞



n=0

(−1) n

p n (a)q n (1)

We prove by induction on n that p n (a) is the sum of n

j=1a i j over all n-tuples

i1, , i n such that i1= 0 and 0 ≤ i j ≤ i j−1+ 1 for 2 ≤ j ≤ n This is precisely our generating function when a k = f (k) for k ≥ 0.

For n = 0, we have the empty product and coefficient 1 For n ≥ 1, let a =

(a1, a2, ) By the induction hypothesis,

∞



n=0

(−1) n

p n (a)q n

1+ a0q

∞



k=0

(−1) k

p k (a)q k = 1,

By examining the coefficient of q non both sides, it follows that

p n (a) = a0

n−1



=0

Group the terms in the claimed formula p n (a) according to the maximum index

j such that i j = 0 By the induction hypothesis, the sum of the terms for which

the last position with subscript 0 in the n-tuple is position l+ 1 can be written as

p  (a)a0p n −1− (a) Hence (2) completes the proof of the formula for p n (a).

The sequence from which we form the continued fraction can be expressed as

a 2r−2 = r and a 2r−1 = 2r for r ≥ 1 Let s = 1/q Our next aim is

1

1+

a0/s

1+

a1/s

1+ · · · =

1 2

∞



k=0

s

s + k

 1 2

k

On the right is the hypergeometric series 12 2F1(s, 1; 1 + s;1

2); in standard notation,

2F1(a, b; c; z) =k≥0 a

(k) b (k)

c (k) z n

n!, where x (k)is the rising factorial We seek a continued

fraction expansion converging to the same value as the left side

It is convenient to multiply both sides of (3) by 1/s The continued fraction on the

left converges when s has positive real part A continued fraction converges to the same

value as its even part, which is the limit of its even approximants (see L Lorentzen and

H Waadeland, Continued Fractions with Applications, North-Holland, 1992, p 84).

The even part is

1/s

1+ a0/s−

a0a1/s2

1+ a1/s + a2/s−

a2a3/s2

1+ a3/s + a4/s− · · ·

With a 2r−2a 2r−1= 2r2and a 2r−1 + a 2r = 3r + 1, this becomes

1/s

1+ 1/s−

2· 12/s2

1+ 4/s−

2· 22/s2

1+ 7/s− · · ·

2· r2/s2

1+ (3r + 1)/s− · · ·

This continued fraction is equivalent, in the sense of having the same sequence of approximants, to

1

s+ 1−

2· 12

s+ 4−

2· 22

s+ 7− · · ·

2· r2

s + 3r + 1− · · ·

To obtain a continued fraction expansion for 2s 21 F1(s, 1; s + 1;1

2), we use the

con-tinued fraction expansion for the standard Euler fraction (see p 308 of the book cited

above) With c= c − a and b= b − a, we obtain

c· 2F1(a, b; c; z)

2F1(a, b + 1; c + 1; z) =

c + (b+ 1)z − (c+ 1)(b + 1)z

c + 1 + (b+ 2)z−

(c+ 2)(b + 2)z

c + 2 + (b+ 3)z−

(c+ 3)(b + 3)z

c + 3 + (b+ 4)z− · · ·

To obtain the desired formula, we set c = a = s, b = 0, and z = 1/2 Since

2F1(s, 0; s;1

2) =k≥0s (k)0(k)

s (k) (1/2) k

k! = 1 and c + r + (b+ r + 1)z = (s + 3r + 1)/2,

the previous display reduces to

s

2F1(s, 1; s + 1;1

2) = s + (1 − s)/2 −

(1)(1)/2 (s + 4)/2−

(2)(2)/2 (s + 7)/2−

(3)(3)/2 (s + 10)/2− · · ·

= 1 2



s+ 1 − 2· 12

s+ 4−

2· 22

s+ 7−

2· 32

s+ 10−



.

Taking the reciprocal of the continued fraction yields

1

2s2F1



s , 1; s + 1,1

2



s+ 1−

2· 12

s+ 4−

2· 22

s+ 7−

2· 32

s+ 10−.

We have thus proved (3)

It remains only to convert 12∞

k=0s +k s

1

2

k

into the generating function we obtained

from the right side of the problem statement Written using q instead of s, the steps

(explained below) are

1

2

∞



k=0

1

1+ kq

 1 2

k

= 1 2

∞



n=0

(−1) n

q n

∞



k=0

 1 2

k

k n

= 1 + 1

2

∞



n=1

(−1) n

q n

∞



k=0

 1 2

kn

m=1

S (n, m)k (m)

= 1 +∞

n=1

(−1) n

q n n



m=1

m !S(n, m) = 1 +∞

n=1

(−1) n

q n n



m=1

m



k=1

(−1) m −k



m k



k n

In step 1 we expand (1 + kq)−1 by the geometric series and interchange the order

of summation In step 2 we invoke k n = S(n, m)k (m) , where S (n, m) is the Stirling

number of the second kind (the number of partitions of n positions into m unordered blocks) and k (m) is the falling factorial Both sides count the k-ary n-tuples, on the right

grouped by the number of distinct entries In step 3 we use the identity∞

k=0

k

m 2−k= 2; this is the instance of ∞

k=0

k

m x k = x m /(1 − x) m+1 at x = 1/2, where the series

converges Finally, in step 4 we use the inclusion-exclusion formula for the Stirling numbers

No solutions were received.

Groups, Rings, Fields, and Power Series

11216 [2006, 366] Proposed by Ted Chinburg, University of Pennsylvania,

Philadel-phia, PA, and Shahriar Shahriari, Pomona College, Claremont, CA Let K be a field,

and let G be an ordered Abelian group The support Supp(a) of a formal sum a =



γ a γ t γ with coefficients a γ in K and exponents γ in G is the set {γ ∈ G : a γ

The generalized power series ring K ((G≤0)) is the set of formal sums a for which

Supp(a) is a well-ordered subset of the nonpositive elements of G Addition and mul-tiplication in K ((G≤0)) are defined in the same way they are for ordinary power

se-ries Show that K ((G≤0)) is Noetherian if and only if either G = {0} or G is

order-isomorphic to Z with the usual ordering (An ordered Abelian group is an Abelian

group G with a total order ≤ such that a ≤ b implies a + c ≤ b + c for all a, b, c

in G.)

Solution by Reid Huntsinger, Lansdale, PA Let R = K ((G≤0

Given a positive elementγ0 in G, let I (γ0) = {a ∈ R : a γ = 0 if γ > −γ0} Always

I (γ0) is an additive subgroup In fact, I (γ0) is an ideal, since a ∈ R and b ∈ I (γ0)

imply(ab) γ =δ≤0 a δ b γ −δ The sum is zero unless γ − δ ≤ −γ0 for someδ ≤ 0,

which requiresγ ≤ −γ0 Thus ab ∈ I (γ0), making I (γ0) an ideal.

On the other hand, if a ∈ R with min Supp(A) = 0 then a is invertible in R (The

inversion algorithm is essentially the customary algorithm for inverting power series with real coefficients with nonzero constant coefficient.) As a result, all proper ideals

of R have the form I (γ ) for some positive γ

Ifγ0 < γ1, then I (γ1) ⊂ I (γ0), and vice-versa Thus the Noetherian property (no

in-finite ascending chains of ideals) is equivalent to having no inin-finite descending chains

in G >0 In other words, R is Noetherian if and only if the positive elements of G are

well-ordered This latter condition means that the positive elements are all powers of

the smallest positive element Thus G is order-isomorphic to Z exactly when R is

Noetherian

Also solved by N Caro (Brazil), R Chapman (U K.), D Fleischman, Szeged Problem Solving Group “Fej´en-tal´altuka” (Hungary), and the proposers.

An Infinite Product Based on a Base

11222 [2006, 459] Proposed by Jonathan Sondow, New York, NY Fix an integer B≥

2, and let s (n) denote the sum of the base-B digits of n Prove that

∞



n=0



k odd

0<k<B

 n B + k

n B + k + 1

(−1) s (n)

= √1

B

Solution for odd B by the GCHQ Problem Solving Group Cheltenham, UK For odd

B we have n ≡ s(n) mod 2 Therefore we can write (−1) n rather than(−1) s (n)in the

infinite product Thus the corresponding infinite series

∞



n=0

(−1) n

log 

k odd

0<k<B

n B + k

n B + k + 1



=∞

n=0

(−1) n+1 

k odd

0<k<B

log



n B + k



converges by the classical alternating series test Hence the product converges to a

finite positive limit By induction (for example), it follows for each positive integer M

that

2M−1

n=0



k odd

0<k<B

 n B + k

n B + k + 1

(−1) n

= (2M B)!2 2M (M!)2

22M B (M B)!(M B)!(2M)!

Using Stirling’s Formula N ! = N N e −N (2π N)1/2 {1 + O(1/N)}, we obtain

2M−1

n=0



k odd

0<k<B

 n B + k

n B + k + 1

(−1) n

= B −1/2 {1 + O(1/M)}.

Also, for n > 0 we have



 log 

k odd

0<k<B

 n B + k

n B + k + 1

(−1) n = 

k odd

0<k<B

log



n B + k



< 

k odd

0<k<B

1

n B + k <

1

n ,

so

2M



n=0



k odd

0<k<B

 n B + k

n B + k + 1

(−1) n

=2M−1

n=0



k odd

0<k<B

 n B + k

n B + k + 1

(−1) n

1+ O(1/M).

Thus for any positive integer N ,

N



n=0



k odd

0<k<B

 n B + k

n B + k + 1

(−1) n

= B −1/2 (1 + O(1/N)) ,

which yields the desired result with a good error estimate

Editorial comment The case B = 2 was the subject of MONTHLYProblem E2692 [85 (1978) 48 and 86 (1979) 394–395].

Other even values of B were treated in a paper by J O Shallit [J Number Theory

21 (1985) 128–134] The case of even B is somewhat more difficult than the case of

odd B, since for even B the exponent (−1) s (n) is a more complicated function of n than (−1) n In particular the familiar alternating series test for establishing convergence of

a series is no longer applicable For a detailed treatment of even B we refer the reader

to Shallit’s paper

Also solved (at least for odd values of B) by K L Bernstein, E J Ionascu, K McInturff, M A Prasad (India),

A E Stadler (Switzerland), A L Stenger, R Tauraso (Italy), and the BSI Problems Group (Germany).

The Square Root of a Certain Matrix

11224 [2006, 459] Proposed by Dietrich Trenkler, University of Osnabr¨uck,

Os-nabr¨uck, Germany, and G¨otz Trenkler, University of Dortmund, Dortmund, Germany.

Let a and b be linearly independent column vectors inR3 Find a formula in terms of a

and b for a square matrix A such that A2= bat− abt (Here, xtdenotes the transpose

of x.)

Solution by Patrick Corn, University of Georgia, Athens, GA Let M = bat− abt

Let

k = a × b (note that k > 0 by linear independence) We compute

M3 = bat (a · b)2− bbt (a · b)(a · a) − ba t (a · a)(b · b)

+ bbt (a · a)(a · b) − aa t (b · b)(a · b) + ab t (a · a)(b · b)

+ aat (a · b)(b · b) − ab t (a · b)2

= (ba t− abt ) (a · b)2− (a · a)(b · b) ,

so

M3+ k2M = M (a · b)2− (a · a)(b · b) + a × b2 = 0.

Letting A = (2k3) −1/2 (M2− kM), we have

A2− M = (2k3)−1(M4− 2kM3+ k2M2− 2k3M )

= (2k3)−1(M − 2k I )(M3+ k2M ) = 0,

as desired (here I is the identity matrix) Writing out A in terms of a and b, we obtain

2a × b3/2

(ba t− abt )2− a × b(ba t− abt )

Editorial comment The restriction to vectors inR3is unnecessary

Also solved by S Amghibech (Canada), C.P Anilkumar (India), M Bataille (France), S.J., Bernau, D R Bridges, R Chapman, (U.K.) P Corn, K Dale (Norway), J.A Grzesik, E.A Herman, K.J Heuvers, R.A Horn, G.L Isaacs, G Keselman, T Kezlan, J.H Lindsey II, O.P Lossers (The Netherlands), P Magli (Italy),

J McHugh, K McInturff, M.D Meyerson, K Onneweer, L Pebody, N.C Singer, A Stadler (Switzerland),

R Stong, J Stuart, T Tam, N Thornber, E.I Verriest, W.C Waterhouse, M Woltermann, J.B Zacharias, BSI Problems Group (Germany) GCHQ Problem Solving Group (U.K.), and the proposer.

A New Bound for r/R

11245 [2006, 760] Proposed by Cezar Lupu, University of Bucharest, Bucharest,

Ro-mania, and Tudorel Lupu, Decebal High School, Constanza, Romania Consider an

acute triangle with sides of lengths a, b, and c, and with an inradius of r and a circum-radius of R Show that

r

R ≤



2(2a2− (b − c)2)(2b2− (c − a)2)(2c2− (a − b)2)

Solution by Ronald A Kopas, Clarion University, Clarion, PA Side a is the sum of

the two segments on either side of the point of tangency with the incircle, so

a = r



cot B

2 + cotC

2



= rcos(B/2) sin(C/2) + cos(C/2) sin(B/2)

sin(B/2) sin(C/2)

= r sin (B/2 + C/2)

sin(B/2) sin(C/2) =

r sin (π/2 − A/2)

sin(B/2) sin(C/2) =

r cos (A/2)

sin(B/2) sin(C/2) ,

and r = a sin(B/2) sin(C/2)/ cos(A/2) From the law of sines, a = 2R sin A = 4R sin (A/2) cos(A/2), so R = a/ 4 sin(A/2) cos(A/2) Therefore

r

R = 4 sin A

2 sin

B

2 sin

C

Now we compute a lower bound for E = 2a2− (b − c)2 We have

E = 2 a2− (b − c)2 + (b − c)2

= 2 b2+ c2− 2bc cos A − b2− c2+ 2bc + (b − c)2

= 4bc (1 − cos A) + (b − c)2 = (b + c)2− (b − c)2 (1 − cos A) + (b − c)2

= (b + c)2(1 − cos A) + (b − c)2cos A

= (b + c)22 sin2 A

2 + (b − c)2cos A ≥ (b + c)22 sin2 A

2,

noting that cos A≥ 0 since the triangle is acute

Therefore √

2 sin(A/2) ≤ 2a2− (b − c)2/(b + c) Similarly √2 sin(B/2) ≤



2b2− (c − a)2/(c + a) and√2 sin(C/2) ≤2c2− (a − b)2/(a + b) Multiply the

product of these three inequalities by√

2 and apply (1) to obtain the required result

Editorial comment In fact the result holds for right and obtuse triangles as well Four

solvers (V Schindler, R Stong, R Tauraso, and the Microsoft Research Problems Group) provided proofs for this stronger result, but their proofs were all computer assisted and much longer than the proof given here

Also solved by A Alt, S Amghibech (Canada), O Bagdasar (Romania), P P D´alyay (Hungary), P De (Ire-land), J Fabrykowski & T Smotzer, V V Garc´ıa (Spain), N Lakshmanan (India), O P Lossers (The Nether-lands), J Minkus, C R Pranesachar (India), T.-L & V R˘adulescu (Romania), J Rooin & A Emami (Iran),

V Schindler (Germany), R Stong, R Tauraso (Italy), D V˘acaru (Romania), L Zhou, Microsoft Research Problems Group, Northwestern University Math Problem Solving Group, and the proposer.

Evaluate the Series

11260 [2006, 939] Proposed by Paolo Perfetti, Mathematics Department, University

“Tor Vergata,” Rome, Italy Find those nonnegative values of α and β for which

∞



n=1

n



k=1

α + k log k

β + (k + 1) log(k + 1)

converges For those values ofα and β, evaluate the sum.

Solution by Richard Stong, Rice University, Houston, TX If α ≥ β, then

∞



n=1

n



k=1

α + k log k

β + (k + 1) log(k + 1) ≥

∞



n=1

α

α + k log k ,

which has all positive terms and diverges So assumeβ > α ≥ 0 Now

n



k=1

α + k log k

β + (k + 1) log(k + 1) =

β

β − α

n



k=1

α + k log k

β + k log k −

n+1



k=1

α + k log k

β + k log k ,

so we have

N−1



n=1

n



k=1

α + k log k

β + (k + 1) log(k + 1) =

α

β − α −

β

β − α

N



k=1

α + k log k

β + k log k .

Since∞

k=1(β − α)/(β + k log k) has all positive terms and diverges, we have

∞



k=1



1− β − α

β + k log k



=∞

k=1

α + k log k

β + k log k = 0.

Hence forβ > α ≥ 0, we have convergence:

∞



n=1

n



k=1

α + k log k

β + (k + 1) log(k + 1) =

α

β − α .

Also solved by O Furdui, J.-P Grivaux (France), O Kouba (Syria), J H Lindsey II, O P Lossers (The Netherlands), A Stadler (Switzerland), Szeged Problem Solving Group “Fej´ental´altuka” (Hungary), GCHQ Problem Solving Group (U K.), Microsoft Research Problems Group, and the proposer.

Group the terms in the claimed formula p n (a) according to the maximum index

j such that i j = By the induction hypothesis, the sum of the. ..

number of the second kind (the number of partitions of n positions into m unordered blocks) and k (m) is the falling factorial Both sides count the k-ary n-tuples, on the right... (γ0) an ideal.

On the other hand, if a ∈ R with Supp(A) = then a is invertible in R (The< /i>

inversion algorithm is essentially the customary algorithm for inverting