THE AMERICAN MATHEMATICAL MONTHLY 4-2008

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 August 31, 2008 Additional information, such as general-izations 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

11355 Proposed by Jeffrey C Lagarias, University of Michigan, Ann Arbor, MI

De-termine for which integers a the Diophantine equation

1

x yz

has infinitely many integer solutions(x, y, z) such that gcd(a, xyz) = 1.

11356 Proposed by Michael Poghosyan, Yerevan State University, Yerevan, Armenia.

Prove that for any positive integer n,

n



k=0

n

k

2

(2k + 1)2n

2k

 = 24n (n!)4

(2n)!(2n + 1)! .

11357 Proposed by Mehmet S¸ahin, Ankara, Turkey Let I a , I b , I c and r a , r b , r c be

respectively the excenters and exradii of the triangle A BC If ρ a,ρ b,ρ care the inradii

of triangles I a BC, I b C A, and I c A B, show that

ρ a

r a

+ρ b

r b

+ ρ c

r c

= 1.

11358 Proposed by Marian Tetiva, National College “Gheorghe Ros¸ca Codreanu”,

Bˆarlad, Romania Let d be a square-free positive integer greater than 1 Show that

there are infinitely many positive integers n such that dn2+ 1 divides n!.

11359 Proposed by J Monterde, University of Valencia, Valencia, Spain Find an

ex-plicit parametric formula for the geometric envelope on the interval(0, 1) of the

fam-ily of Bernstein polynomials B n

s (t), defined by B n

s (t) =n

s



t s (1 − t) n −s for s ∈ [0, n],

wheren

s



(s+1)(n−s+1).

11360 Proposed by Cezar Lupu, student, University of Bucharest, Bucharest, and

Tu-dorel Lupu, Decebal High School, Constanta, Romania Let f and g be continuous

real-valued functions on[0, 1] satisfying the condition1

0 f (x)g(x) dx = 0 Show that

1

0 f21

2

and1

2

+1

2

≥ 41

2

11361 Proposed by Finbarr Holland, University College Cork, Cork, Ireland The

Lemoine point of a triangle is the unique point L inside the triangle such that the

distances from L to the sides are proportional to the corresponding side lengths Given a circle G and distinct fixed points B, C on G, let K be the locus of the Lemoine point of A BC as A traverses the circle Show that K is an ellipse.

SOLUTIONS

A Vanishing Alternating Sum

11212/11220 [2006, 268/367] Proposed by David Beckwith, Sag Harbor, NY Show

that when n is a positive integer,

n



r=0

(−1) r



n r



2n − 2r

= 0.

Solution I by Harris Kwong, SUNY, Fredonia, NY From (1 − t) −n = ∞k=0

k +n−1

n−1



t k

and(1 − t2) n= n

r=0(−1) rn

r



t 2r, we obtain

(1 + t) n = (1 − t2) n

(1 − t) n =∞

k=0

n



r=0

(−1) r



n r



k + n − 1

t k +2r

Comparing coefficients of t n+1yields the desired identity.

Solution II by Rob Pratt, Raleigh, NC More generally, if k < n, then

n



r=0

(−1) r



n r



2n − 2r

k

= 0.

Since k < n, no k-person committees can be formed from n couples using at least one

member of each couple To count these committees using inclusion-exclusion, let A i

be the set of k-person committees using neither member of couple i Now i ∈S A i =

2n −2|S|

k



Hence the sum on the left is the inclusion-exclusion sum to count the empty set

Solution III by Nicholas Singer, Annandale, VA Indeed, n r=0(−1) rn

r



P (r) = 0 for

any polynomial P of degree less than n The nth forward difference of the sequence

(P(0), P(1), ) has first term (−1) n n

r=0(−1) rn

r



P (r) Since P is a polynomial of

degree less than n, its nth forward difference is identically zero.

Also solved by 62 other readers and the proposer.

Sometimes an Integer

11213 [2006, 268] Proposed by Stanley Rabinowitz, Chelmsford, MA For positive

integers n and m with n odd and greater than 1, let S (n, m) = (n−1)/2 k=1 sec2m ( k π

n+1).

(a) Show that if n is one less than a power of 2, then S (n, m) is a positive integer.

(b∗) Show that if n does not have the form of part (a), then S (n, m) is not an integer.

Solution to (a) by NSA Problems Group, Fort Meade, MD Let n+ 1 = 2s with s≥ 2.

We use induction on s The base case s= 2 is easy: sec2m (π/4) = 2 m Note that

S (2 s+1− 1, m) =

2s−1

k=1

sec2m



k π

2s+1

= S(2 s − 1, m) +

2s−1



j=1

sec2m



(2 j − 1)π

2s+1

.

By the induction hypothesis, it suffices to show that the last sum is integral

The Chebyshev polynomial T r is a polynomial of degree r defined by T r (x) =

cos(r arccos(x)) In particular, T2s is a polynomial of degree 2s with T2s (x) = 0 when

x = cos((2 j − 1)π/2 s+1) for 1 ≤ j ≤ 2 s Furthermore, since T2(x) = 2x2− 1 and

T2s (x) = 2 cos2(2 s−1arccos(x)) − 1 = 2T2s−1(x)2− 1,

it follows that T2s (x) ∈ Z[x] and, apart from the constant term, all of its

coeffi-cients are even Since T2s (0) = cos(2 s π/2) = 1 for s ≥ 2, the reciprocal polynomial

x2s T2s (1/x) is a monic integral polynomial that has a zero at each of the 2 S numbers

x k := sec((2 j − 1)π/2 s+1), 1 ≤ j ≤ 2 s Furthermore, every elementary symmetric function in 2Svariables evaluates to an even integer at(x1, , x2S ) Now

2s−1



j=1

sec2m



(2 j − 1)π

2s+1

= 1 2

2s



j=1

sec2m



(2 j − 1)π

2s+1

= 1 2

2S



j=1

x 2m j ,

and from the foregoing observations, this is an integer, as desired

Editorial comment Unfortunately, due to an error by the editors, part (b∗) of this

prob-lem was incorrectly stated It should have read, “Show that if n does not have the form

of part (a), then there exists a positive integer m such that S (n, m) is not an integer.”

Several solvers noticed that part (b∗) as stated is false, with a simple counterexample

being S (9, 1) = 16 The intended statement of part (b∗) remains untreated.

Also solved by D R Bridges, R Chapman (U K.), P P Dalyay (Hungary), Y Dumont (France), J W From-meyer, J Grivaux (France), G Keselman, O P Lossers (The Netherlands), M A Prasad (India), A Stadler (Switzerland), A Stenger, R Stong, M Tetiva (Romania), A Tissier (France), BSI Problems Group (Ger-many), GCHQ Problem Solving Group (U K.), and the proposer.

The Number Between 1 and n That is Least Prime

11218 [2006, 366] Proposed by Gary Gordon, Lafayette College Consider the

follow-ing algorithm, which takes as input a positive integer n and proceeds by rounds, listfollow-ing

in each round certain positive integers between 1 and n inclusive, ultimately producing

as output a positive integer f (n), the last number to be listed In the 0th round, list 1.

In the first round, list, in increasing order, all primes less than n In the second round,

list in increasing order all numbers that have not yet been listed and are of the form

2 p, where p is prime Continue in this fashion, listing numbers of the form 3 p, 4 p, and so on until all numbers between 1 and n have been listed Thus f (10) = 8 because

the list eventually reaches the state(1, 2, 3, 5, 7, 4, 6, 10, 9, 8), while f (20) = 16 and

f (30) = 27.

(a) Find f (2006).

(b) Describe the range of f

(c) Find limn→∞f (n)/n and lim n→∞f (n)/n.

Solution by Bruce S Burdick, Roger Williams University, Bristol, RI In describing the

first round of the process, “less than n” should be “at most n”.

(b) Let R be the range of f , and let S = {2i3j : i is a nonnegative integer and j = 0

or 3· 2l−1≤ 3j < 2 l+1for some l ∈ N} We show that R = S.

Let g (1) = 0 For k ≥ 2, let g(k) = k/p, where p is the largest prime divisor of k.

When k ≤ n, the number k appears in round g(k) Thus f (n) is the largest element of [n] that maximize g.

Let T = {k ∈ N: g(k) ≥ g(i) for 1 ≤ i < k} If k ∈ T , then f (k) = k; hence

T ⊆ R Furthermore, f (n) = k requires k ∈ T , so R = T To compare T and S, we show first that a number in T has no prime divisor outside {2, 3} If g(k) = k/p with

p > 4, then there is a positive integer m such that 2k/p < 2 m < k Now g(2 m ) =

2m−1 > k/p = g(k) and k /∈ T

Now consider the exponents Note that T contains all powers of 2, since g (2 m ) =

2m−1 > g(k) for k < 2 m If 2i3j j−1 ≥ 2l−1whenever 2l < 3 j; otherwise 2i +l < 2 i3j and g (2 i3j ) < g(2 i +l ) Letting 2 l be the largest power of 2 less than 3j, we obtain 3· 2l−1≤ 3j < 2 l+1 Hence T ⊆ S Finally, if 2 i3j ∈ S, then 2 i3jis listed after all smaller numbers divisible by 3 and after all smaller powers of 2 Hence

S ⊆ R, and we have R = S.

The condition for 2i3j

binary expansion of 3j is 1 It is also equivalent to the fractional part of j log23 being

at least log23 The first few values of j with this property are 1, 3, 5, 8, 10, 13, 15, 18,

and 20

(a) f (2006) = 1944 To find f (2006) using (b), we create a list of candidates by

taking powers of 2 and numbers of the form 2i3j where j ∈ {1, 3, 5} and 2 i is the largest power of 2 bounded by 2006/3 j The candidates are 1024, 1536, 1728, and

1944 The largest, 1944, is f (2006).

(c) limn→∞f (n)/n = 1 When n ∈ R, the value of f (n)/n is 1 There are infinitely

many such n, and f (n) ≤ n for all n, so lim n→∞f (n)/n = 1.

limn→∞f (n)/n = 2/3 When n = 3 · 2 i − 1, the value of f (n) is 2 i+1, so f (n)/n =

2i+1/(3 · 2 i − 1) = 2/(3 − 2 −i ) These numbers tend to 2/3 If 2 i+1≤ n ≤ 3 · 2 i− 1,

then f (n) = 2 i+1, and f (n)/n > 2/3 Similarly, if 3 · 2 i ≤ n ≤ 2 i+2− 1, then f (n) ≥

3· 2i , and f (n)/n > 3/4 Hence lim n→∞f (n)/n = 2/3.

Solved also by M R Avidon, D Beckwith, P Corn, D Cranston, D Fleischman, J W Frommeyer, J.-P Gri-vaux (France), C C Heckman, E A Herman, M Hildebrand, M Huibregtse, G Keselman, J H Lindsey

II, O P Lossers (The Netherlands), D Lovit, F B Miles, D Opitz, W Y Pong, M A Prasad (India),

A Stadler (Switzerland), R Staum, J H Steelman, R Stong, M Tetiva (Romania), L Zhou, BSI Problem Solving Group (Germany), Szeged Problem Solving Group “Fej´ental´altuka” (Hungary), GCHQ Problem Solv-ing Group (U K.), Manchester Problem SolvSolv-ing Group (U K.), Microsoft Research Problems Group, and Northwestern University Problem Solving Group.

A Romanian Olympiad Problem Generalized

11248 [2006, 760] Proposed by P´al P´eter D´alyay, De´ak Ferenc High School, Szeged,

Hungary Let n be a positive integer, and let f be a continuous real-valued

func-tion on[0, 1] with the property that1

0 x k f (x) dx = 1 for 0 ≤ k ≤ n − 1 Prove that

1

0( f (x))2d x ≥ n2 (The case n= 2, due to Ioan Ras¸a, appeared on the 55th

Roma-nian Mathematical Olympiad, and the solution was published in Gazeta Matematic˘a

(CIX) 5-6 (2004), page 227.)

Solution by Jaime Vinuesa, University of Cantabria, Santander, Spain The

clas-sical Legendre polynomial P k is a polynomial of degree k given by 2 k k !P k (x) =

(d k /dx k )[(x2− 1) k ] These polynomials satisfy P k (1) = 1 and

1

−1

P j (x)P k (x) dx =

⎧

⎪

⎪

2

2k+ 1 if j = k.

Let Q k (x) =√2k + 1 P k (2x − 1) Now {Q0, Q1, , Q n−1} is an orthonormal set

in L2[0, 1] Moreover, f, Q k

1

0 f (x)Q k (x) dx = Q k (1) =√2k+ 1, so Bessel’s inequality yields

1

0

( f (x))2d x =  f 2≥

n−1



k=0

| f, Q k 2 =

n−1



k=0

(2k + 1) = n2.

Editorial comment As some contributors noted, the inequality is sharp, and it holds

for all f ∈ L2[0, 1] Some solvers observed that it suffices to construct a polynomial p

of degree at most n− 1 satisfying1

0(p(x))2d x = 1 and p(1) = n The desired result

then follows from the Schwarz inequality using

orthogonal polynomials, Lagrange interpolation, Hilbert matrices—were used in con-structing such a polynomial

Also solved by U Abel (Germany), O J L Alfonso (Colombia), S Amghibech (Canada), K F Andersen (Canada), A Bandeira & E Dias (Portugal), M Benito & ´ O Benito & E Fern´andez (Spain), M W Botsko,

P Bracken, D R Bridges, M A Carlton, R Chapman (U K.), A Chaudhuri & R Selukar, P R Chernoff,

K Dale (Norway), Y Dumont (France), B Dunn III, J Fabrykowski & T Smotzer, S Foucart, J.-P Grivaux (France), J Groah, E A Herman, F Holland (Ireland), E J Ionascu (Romania), M E H Ismail, D Jespersen,

J Kalis, G Keselman, O Kouba (Syria), A V Kumchev, K.-W Lau (China), J H Lindsey II, O P Lossers (The Netherlands), R Martin (U K.), W Matysiak (Poland), T Nowak (Austria), M A Prasad (India), H Ricardo, J Rooin & M Hassani (Iran), H.-J Seiffert (Germany), J G Simmonds, N C Singer, A Stadler (Switzerland), V Stakhovsky, A Stenger, R Stong, J Sun, M Tetiva (Romania), S Vagi, Z V¨or¨os (Hun-gary), R Whitley, L Zhou, BSI Problems Group (Germany), Szeged Problem Solving Group “Fej´ental´altuka” (Hungary), GCHQ Problem Solving Group (U K.), Microsoft Research Problems Group, and the proposer.

A Symmetric Inequality

11255 [2006, 848] Proposed by Slavko Simic, Mathematical Institute SANU,

Bel-grade, Serbia Let n be a positive integer, x1, , x n be real numbers, and let

i=1 p i x k

i −

n

i=1 p i x i

k

Show that for m ≥ 1,

S 2m S 2m+2≥



S 2m+12 .

Solution by O P Lossers, Eindhoven University of Technology, Eindhoven, The Netherlands We assume that p1, , p n are nonnegative real numbers summing

to 1, because if p = (1, −3, 3) and x = (1, 2, 3), then S2S4 < (2/3)S2

3 and the in-equality fails

Define f: R2→ R by

2t x 2m+1

x 2m

2m (2m − 1) .

Now ∂x ∂22 f (x, t) = (tx m + x m−1)2≥ 0, so f (x, t) is a convex function of x for all t.

From Jensen’s inequality it then follows that, for all t,

n



i=1

p i f (x i , t) − f

 n



i=1

p i x i , t



≥ 0,

or, equivalently,

t2S 2m+2

2t S 2m+1

S 2m

The expression on the left, considered as a polynomial in t, must therefore have a

nonpositive discriminant:



2S 2m+1

2

− 4 S 2m+2

S 2m

and this is equivalent to the desired inequality

Editorial comment The requirement that p i ≥ 0 was given by the proposer but lost

by the editors A more subtle oversight is the unstated requirement that m should be

an integer Byron Schmuland provided a counterexample with m = 3/2, x1= x2= 1,

x3 = −1, and p1 = p2= p3= 1/3, in which case S3S5< (5/6)S2

4

Also solved by B Schmuland (Canada), Microsoft Research Problems Group, and the proposer.

Perimeter by Integral

11256 [2006, 848] Proposed by Finbarr Holland, University College Cork, Cork,

Ire-land For complex a, b, and c, let f (x) = max{Re(ae i x ), Re(be i x ), Re(ce i x )} Find

2π

Solution by Michel Bataille, Rouen, France Let I =2π

|a − b| + |b − c| + |c − a|; this is the perimeter of the triangle formed by the three

complex numbers

Let g (x) = min{Re(ae i x ), Re(be i x ), Re(ce i x )} Now

2π

0

0

max{−Re(ae i x ), −Re(be i x ), −Re(ce i x )} dx

= − 2π

0

max{Re(ae i (x+π) ), Re(be i (x+π) ), Re(ce i (x+π) )} dx

= − 3π

π

max{Re(ae i t ), Re(be i t ), Re(ce i t )} dt = −I,

since f is 2 π-periodic Now max{u, v} = 1

2π

0 e i x d x = 0, so with the notation J a ,b=2π

0 max{Re(ae i x ), Re(be i x )} dx it follows

that

J a ,b= 1

2Re

2π

0

(ae i x + be i x ) dx

+1 2

2π

0

|Re(ae i x ) − Re(be i x )| dx

= 1

2

2π

0

|Re(ae i x ) − Re(be i x )| dx.

Now let a = α + iαand b = β + iβwhereα, α,β, and βare real numbers Then

Re(ae i x ) − Re(be i x ) = (α − β) cos x − (α− β) sin x = A sin(x + φ),

where A = ((α − β)2+ (α− β)2)1/2 = |a − b|, sin φ = (α − β)/A, and cos φ =

−(α− β)/A Thus,

2π

0

[Re(ae i x ) − Re(be i x )| dx = |a − b| 2π

0

| sin(x + φ)| dx

= |a − b| 2π

0

| sin x| dx = 4|a − b|,

and J a ,b = 2|a − b| Similarly, with K a ,b=2π

0 min{Re(ae i x ), Re(be i x )} dx, we have

K a ,b = −2|a − b| Finally, for real numbers u, v, and w, we have

max{u, v, w} − min{u, v, w} =1

− min{u, v} − min{v, w} − min{w, u}),

so

2I = 2π

0

( f (x) − g(x)) dx = 1

2(J a ,b + J b ,c + J c ,a − K a ,b − K b ,c − K c ,a )

= 1

2



2|a − b| + 2|b − c| + 2|c − a| + 2|a − b| + 2|b − c| + 2|c − a|

= 2|a − b| + 2|b − c| + 2|c − a|.

This is the stated result

Also solved by S Amghibech (Canada), M R Avidon, D R Bridges, B S Burdick, R Chapman (U K.),

A Demis (Greece), J.-P Grivaux (France), E A Herman, F Hjouj, G Keselman, J H Lindsey II, M D Meyerson, J G Simmonds, V Stakhovsky, R Stong, R Tauraso (Italy), E I Verriest, J Vinuesa (Spain),

H Widmer (Switzerland), J H Zacharias, L Zhou, Szeged Problem Solving Group “Fej´ental´altuka” (Hun-gary), GCHQ Problem Solving Group (U K.), Microsoft Research Problems Group, and the proposer.

A Choice Sum

11263 [2006, 940] Proposed by Gregory Keselman, Oak Park, MI, and formerly of

Lvov Polytechnic Institute, Ukraine Show that when n is a positive integer and a is

real,

n/2

k=0

(−1) k

a k



k

=

⎧

⎪

⎪

(1+√1−4a) n+1−(1−√1−4a) n+1

2n+1 √

a n /2 cos(nβ) + sin(nβ)/√4a− 1, if a > 1/4.

Here,β denotes arcsin√1− 1/(4a).

Solution by E Leonard, University of Alberta, Edmonton, AL, Canada For each

non-negative integer n, define

b n=n/2

k=0

(−1) k

a k



k

=

n



k=0

(−1) k

a k



k

.

r



=m

r

 +m

r−1

 andm

r



= 0 if r > m, we see that the sequence b n satisfies

This linear recurrence has characteristic equationλ2− λ + a = 0, which has roots:

2±1 2

√

4,

4,

2± i1

2

√

4.

(i) For a < 1/4 there are two distinct roots, so the general solution of the recurrence

is

b n = A



1+√1− 4a

2

n

+ B



1−√1− 4a

2

n

.

We now compute A and B from the initial conditions b0 = b1= 1 to arrive at

b n =



1+√1− 4an+1

−1−√1− 4an+1

2n+1√

(ii) For a = 1/4 there is a double root, so the general solution of the recurrence is

2n + Bn

2n

Again using the initial conditions to determine A and B, we obtain

b n = n+ 1

2n

(iii) For a > 1/4, there are two distinct complex roots, which may be written in

polar form as

√

a

cosβ + i sin β=√ae i β , and √a

cosβ − i sin β=√ae −iβ ,

whereβ = arcsin√1− 1/(4a) The general solution of the recurrence is

b n = Aa n /2 e i n β + Ba n /2 e −inβ

Applying the initial conditions to evaluate A and B, and using

sinβ =

√

4a− 1

2√

2√

a ,

we get

b n = a n /2 cos n β + a

n /2

√

4a− 1sin n β.

Also solved by U Abel (Germany), S Amghibech (Canada), M R Avidon, M Bataille (France), D Beckwith,

J Borwein (Canada), P Bracken, R Chapman (U.K.), P P D´alyay (Hungary), P De (Ireland), G.C Greubel, J.-P Grivaux (France), H Kwong, G Lamb, O P Lossers (The Netherlands), J Minkus, C R Pranesachar (India), T L & V Rˆadulescu (Romania), H Roelants (Belgium), J N Senadheera, A Stadler (Switzerland),

R Stong, M Tetiva (Romania), H Widmer (Switzerland), C Zhou (Canada), BSI Problems Group (Germany), GCHQ Problem Solving Group (U.K.), Microsoft Research Problems Group, and the proposer.

Editorial comment The requirement that p i ≥ was given by the proposer but lost

by the editors A more subtle oversight is the. .. 1/4 there is a double root, so the general solution of the recurrence is

2n + Bn

2n

Again using the. .. Eindhoven, The Netherlands We assume that p1, , p n are nonnegative real numbers summing

to 1, because if p = (1, −3, 3) and x = (1, 2, 3), then S2S4