Tạp chí toán học AMM tháng 6 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 October 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

11369 Proposed by Donald Knuth, Stanford University, Stanford, CA Prove that for

all real t, and all α ≥ 2,

e αt + e −αt − 2 ≤e t + e −tα

− 2α

11370 Proposed by Michael Goldenberg and Mark Kaplan, Baltimore Polytechnic

In-stitute, Baltimore, MD Let A0, A1, and A2be the vertices of a non-equilateral triangle

T Let G and H be the centroid and orthocenter of T , respectively Treating all indices modulo 3, let B k be the midpoint of A k−1A k+1, let C kbe the foot of the altitude from

A k , and let D k be the midpoint of A k H

The nine-point circle of T is the circle through all B k , C k , and D k We now introduce nine more points, each obtained by intersecting a pair of lines (The intersection is not

claimed to occur between the two points specifying a line.) Let P kbe the intersection

of B k−1C k+1 and B k+1C k−1, Q k the intersection of C k−1D k+1 and C k+1D k−1, and R k the intersection of C k−1C k+1 and D k−1D k+1.

Let e be the line through {P0, P1, P2}, and f be the line through {Q0, Q1, Q2}

(By Pascal’s theorem, these triples of points are collinear.) Let g be the line through {R0, R1, R2}; by Desargues’ theorem, these points are also collinear

(a) Show that the line e is the Euler line of T , that is, the line through the circumcenter

and centroid of T

(b) Show that g coincides with f

(c) Show that f is perpendicular to e.

(d) Show that the intersection S of e and f is the inverse of H with respect to the

nine-point circle

11371 Proposed by Ovidiu Furdui, University of Toledo, Toledo, OH Let A denote

the Glaisher-Kinkelin constant, given by

A= lim

n→∞n

−n2/2−n/2−1/12 e n2/4n

k=1

k k = 1.2824 · · ·

Evaluate in closed form

A6

∞



n=1



e−1(1 + 1/n) n(−1) n

.

11372 Proposed by Jennifer Vandenbussche and Douglas B West, University of

Illi-nois at Urbana-Champaign, Urbana, IL In a family of finite sets, let a splitting ele-ment be an eleele-ment that belongs to at least two of the sets and is omitted by at least

two of the sets Determine the maximum size of a family of subsets of{1, , n} for

which there is no splitting element

11373 Proposed by Emeric Deutsch, Polytechnic University, Brooklyn, NY Let S n

be the symmetric group on {1, , n} By the canonical cycle decomposition of an

elementπ of S, we mean the cycle decomposition of π in which the largest entry of

each cycle is at the beginning of that cycle, and the cycles are arranged in increasing order of their first elements

Let ψ n : S n → S n be the mapping that associates to each π ∈ S n the permuta-tion whose one-line representapermuta-tion is obtained by removing the parentheses from the canonical cycle decomposition ofπ (Thus, the permutation12345

34521

 has one-line repre-sentation 34521 and canonical cycle reprerepre-sentation(42)(513) and is mapped by ψ5to 42513.) Describe the fixed points ofψ nand find their number

11374 Proposed by Harley Flanders and Hugh L Montgomery, University of

Michi-gan, Ann Arbor, MI Let a, b, c, and m be positive integers such that abcm = 1 + a2+

b2+ c2 Show that m= 4

11375 Proposed by Cezar Lupu, student, University of Bucharest, Bucharest,

Roma-nia The first Brocard point of a triangle A BC is that interior point  for which the

anglesBC, C A, and AB have the same radian measure Let ω be that measure.

Regarding the triangle as a figure in the Euclidean planeR2, show that if the vertices belong toZ × Z, then ω/π is irrational.

SOLUTIONS

Consecutive Squares with a Square Average

11227 [2006, 460] Proposed by Syrous Marivani, Louisiana State University –

Alexandria, Alexandria, LA Call an integer n blocky if n > 1 and there is a run

of n consecutive integer squares the average of which is a square (Thus, 25 is blocky

because(24

0 k2)/25 = 4900/25 = 142, and 31 is the next blocky integer.)

(a) Determine the set B of blocky integers.

(b) Given a blocky integer n, give a procedure that determines all integers k that can

serve as starting points for required runs of squares

(c) Give a formula in terms of n for the number of k that can serve as starting points

of required runs of squares

Solution by Missouri State University Problem Solving Group, Springfield, MO.

(a) We prove that n is blocky if and only if n > 1 and n ≡ ±1 or ±7 (mod 24) We want to find integers k, n, and m such that

m2= 1

n

k +n−1

i =k

i2 = 1

n

k +n−1



i=1

i2−

k−1



i=1

i2



.

Using the standard formulaN

i=1i2 = N(N + 1)(2N + 1)/6 and algebraically simpli-fying, we obtain m2= k2+ (n − 1)k + (n − 1)(2n − 1)/6 Thus (n − 1)(2n − 1)/6 must be an integer, which forces n ≡ ±1 (mod 6) Completing the square on the right side yields m2= (k + (n − 1)/2)2+ (n2− 1)/12, so

n2− 1

12 = m2−

k+n− 1 2

2

=

m + k + n− 1

2 m − k − n− 1

Since n = 6l ± 1, (n − 1)/2 = 3l or 3l − 1, while (n2− 1)/12 = 3l2± l, which is

an even integer Because m + k + (n − 1)/2 and m − k − (n − 1)/2 have the same

parity and their product is even, we have(n2− 1)/12 ≡ 0 mod 4 This forces n ≡ ±1

or±7 (mod 24) Conversely, if n is of this form, then (n2− 1)/12 = (2α)(2β), where

α and β are integers Taking m = α + β and k = α − β − (n − 1)/2 proves that the

conditions are sufficient

(b) Given a blocky integer n, taking all possible factorizations of the form (n2−

1)/12 = (2α)(2β) with α, β > 0 and setting k = α − β − (n − 1)/2 yields all pos-sible starting points k for the summation (Notice that if α = −a and β = −b with

a , b > 0, then k also arises from the positive factorization (n2− 1)/12 = (2a)(2b), because k = α − β − (n − 1)/2 = b − a − (n − 1)/2.)

(c) From the construction above, we conclude that the number of such k is the

number of positive divisors of(n2− 1)/48, which is easily computed from its prime

factorization

Also solved by K Alpar-Vajk (Italy), M R Avidon, D Beckwith, R Chapman (U K.), P Corn, P P D´alyay (Hungary), J Ferdinands, J W Frommeyer, S M Gagola, Jr., E A Herman, O P Lossers (Netherlands),

P Magli (Italy), L Pebody, F Sami, R Stong, Z V¨or¨os (Hungary), Con Amore Problem Group (Denmark), GCHQ Problem Solving Group (U K.), Microsoft Research Problems Group, NSA Problems Group, and the proposer.

Inscribed Triangle Area

11261 [2006, 940] Proposed by Isaac Sofair, Fredericksburg, VA A triangle of area

1 has vertices A1, A2, and A3 The sides A2A3, A3A1, and A1A2 subtend angles of measureα1,α2, andα3, respectively, at an internal point P The triangle has angles at

A1, A2, and A3of measure a1, a2, and a3, respectively The extensions of A1P, A2P, and A3P to their opposite sides meet those sides at B1, B2, and B3, respectively For 1≤ k ≤ 3, let T k = sin a ksinα ksin(α k − a k ) For an even permutation p of (1, 2, 3), let S p = (sin a p1sinα p2sinα p3 + sin α p1sin a p2sin a p3), and let S be the product of S p over all three such permutations Prove that the area of triangle B1B2B3

is 2T1T2T3/S.

Solution by C R Pranesachar, Indian Institute of Science, Bangalore, India Notation:

We write, for example, A2B1 for the length of segment A2B1, and[P A2A3] for the area of P A2A3 If t1= A2B1/B1A3, t2 = A3B2/B2A1, and t3 = A1B3/B3A2, then

t1t2t3= 1 by Ceva’s theorem (see

[A1B3B2] = 1

2 A1B3· A1B2sin a1= 1

2

t3

1+ t3

A1A2· 1

1+ t2

A1A3sin a1

(1 + t3)(1 + t2) [A1A2A3].

From this and the two similar relations for[A2B1B3] and [A3B2B1] we get

[B1B2B3] = [A1A2A3] − [A1B3B2] − [A2B1B3] − [A3B2B1]

(1 + t1)(1 + t2)(1 + t3) [A1A2A3].

Subtracting areas, we have

[A1A2B1]

[A1B1A3] = t1, [P A2B1]

[P B1A3] = t1, so [P A1A2]

[P A3A1] = t1, and similarly for t2, t3

Now write

a1= x1+ x2

a2= y1+ y2

a3= z1+ z2

dividing the angles

us-ing lines through P as

shown in the diagram

A2

A1

B1

A3

B3

B2

y2

2

z1

α1

α2

α3

x1

x2

P

We obtain

sin x1 sin z2 = (A2P /A1A2) sin α3

(A2P /A2A3) sin α1

= sinα3sin a1

sinα1sin a3 Also x1+ z2 = A1P A3− A1A2A3 = α2− a2 Observe that if sinθ/ sin φ = k and

θ + φ = β, then

tanθ = k sin β

1+ k cos β and tanφ =

sinβ

k + cos β .

Therefore

tan z2= sin a3sinα1sin(α2− a2)

sin a1sinα3+ sin a3sinα1cos(α2− a2) .

Similarly (interchanging suffixes 2 and 3)

tan y1 = sin a2sinα1sin(α3− a3)

sin a1sinα2+ sin a2sinα1cos(α3− a3) .

Hence

[P A2A3] = 1

2(A2A3)2 tan y1tan z2

tan y2+ tan z2

= (A2A3)2

2

sin a2sin a3sinα1sin(α2− a2) sin(α3− a3)

sin a3sin(α2− a2) {sin a1sinα2+ sin a2sinα1cos(α3− a3)}

+ sin a1sin(α3− a3) {sin a1sinα3+ sin a3sinα1cos(α2− a2)}

= [A1A2A3] sin a1sinα1sin(α2− a2) sin(α3− a3)

sin a p1sin a p2sinα p3sin(α p3 − a p3) ,

is the sum over the even permutations Writing

x = sin a1sinα1sin(α2− a2) sin(α3− a3),

y = sin a2sinα2sin(α3− a3) sin(α1− a1),

z = sin a3sinα3sin(α1− a1) sin(α2− a2),

we have t1 = z/y, t2= x/z, t3 = y/x So

x yz = T1T2T3sin(α1− a1) sin(α2− a2) sin(α3− a3),

y + z = sin(α1− a1) [sin a2sinα2sin(α3− a3) + sin a3sinα3sin(α2− a2)]

Expanding and rearranging gives

y + z = sin(α1− a2) [sin a2sin a3sinα1+ sin a1sinα2sinα3]= sin(α1− a1) S (123) Using this and similar formulas for z + x and x + y, we get

[B1B2B3] = 2T1T2T3

S [A1A2A3],

as desired

Also solved by V Schindler (Germany) and the proposer.

A Combinatorial Inequality

11269 [2007, 78] Proposed by Herv´e Moulin, Rice University, Houston, TX Let n be a

positive integer and let a1, , a n be nonnegative real numbers For integers 1≤ k ≤

n, and for p ≥ 1, let A(k, p) =S



j ∈S a j

p

, where the outer summation extends over all subsets of{1, , n} having exactly k elements Thus, A(n, p) =n

j=1a j

p

while A (1, p) =n

j=1a

p

j

(a) Show that if p ≥ n − 1 or p is a positive integer, then

A (n, p)

n−1



k=0

(−1) k A (n − k, p)

(b) Show that if p ≥ n or p is a positive integer, then

n−1



k=0

(−1) k A (n − k, p)

1

n+ 1

n+ 1

n

p

A (n, p).

(c)∗Does the conclusion of part (a) hold also for nonintegral p with 1 < p < n − 1? Solution by the editors (a) More generally, let f be a function differentiable n times, such that f (x), , f (n) (x) ≥ 0 on (0, ∞) and f (0) = 0 For any subset S of [n] let

s ∈S

a s , b S =

T ⊆S

(−1) |S|−|T | f (a T ).

Note that for nonempty S, b S is an |S|-fold iterated finite difference of f By the

mean value theorem there is someξ ∈ (0,s ∈S a s ) with b S = s ∈S a s



f (|S|) (ξ) In particular, b S ≥ 0 for nonempty S, and b∅= 0 Also note that by inclusion-exclusion,

f (a S ) =

T ⊆S

b T



S ⊆[n]

(−1) n −|S|

n + 1 − |S| f (a S ) = 

S ⊆[n]

(−1) n −|S|

n + 1 − |S|



T ⊆S

T ⊆[n]

b T



S ⊃T

(−1) n −|S|

n + 1 − |S|

T ⊆[n]

b T

n−|T |

k=0

(−1) k

k+ 1

n − |T |

T ⊆[n]

b T

n + 1 − |T |

≥ 1

n



T ⊆[n]

b T = 1

n f

 n



j=1

a j



.

Here in the third equality we let k = n − |S| For the inequality we have used b T ≥ 0

for nonempty T , and b∅ = 0 For part (a) we take f (x) = x p, which has the desired

nonnegativity property exactly when p ≥ n − 1 or p is a positive integer For use in

part (b), note that similarly



S ⊆[n]

(−1) n −|S|

n + 2 − |S| f (a S ) = 

T ⊆[n]

b T

n−|T |

k=0

(−1) k

k+ 2

n − |T | k

T ⊆[n]

b T (n + 1 − |T |)(n + 2 − |T |) ≥ 0 (b) We show that if f is an n + 1-times differentiable function such that f (0) = 0 and f (x), , f (n+1) (x) ≥ 0 on (0, ∞), then



S ⊆[n]

(−1) n −|S|

n + 1 − |S| f (a S ) ≤ 1

n+ 1f



n+ 1

n

n



j=1

a j



.

The desired result is then the special case f (x) = x p Let F (a1, , a n ) denote the left side of this desired inequality and suppose that a n > a n−1 Since

∂ F

∂a k

S :k∈S

(−1) n −|S|

n + 1 − |S| f(a S ),

we have

∂ F

∂a n

− ∂ F

∂a n−1 = − 

S ⊆[n−2]

(−1) n −2−|S|

(n − 2) + 2 − |S|



f(a n + a S ) − f(a n−1+ a S )≤ 0.

Here for the inequality we have used the result of the previous paragraph for the

func-tion g given by g (x) = f(a n + x) − f(a n−1+ x) Since the symmetric result holds

for the other variables, it follows that for fixedn

j=1a j , F attains its maximum when

a1 = · · · = a n Thus it suffices to prove the inequality when a1= · · · = a n = a If we take a1 = · · · = a n = a n+1 = a then

b [n+1]=

n+1



k=0

(−1) n +1−k

n+ 1

k f (ka) = f ((n + 1)a) −

n



k=0

(−1) n −k (n + 1)

n + 1 − k f (ka). Since b [n+1]≥ 0,

n



k=0

(−1) n −k

n + 1 − k f (ka) ≤

1

n+ 1f ((n + 1)a).

This is exactly the desired inequality in the case where a1= · · · = a n = a Note that if

f (x) = x p with n − 1 < p < n, then f (n+1) (x) < 0 on (0, ∞) Hence b [n+1] < 0 and the reverse inequality holds Thus (b) does not hold for n − 1 < p < n.

Editorial comment No solution of (c)∗is available

Parts (a) and (b) were also solved by the proposer.

The Euler Line After All

11272 [2007, 164] Proposed by Vasile Mihai, Belleville, ON, Canada Let A BC be

an acute nonequilateral triangle, and let H be its orthocenter, O its circumcenter, and

K its symmedian point (defined below) Let R be the circumradius of A BC Let L be the line through H and parallel to the line segment K O.

(a) Show that there are exactly two solution points V on L to the equations

|V A|

|BC| =

|V B|

|C A| =

|V C|

(b) For the solutions V1and V2in (a), show that|H V1| · |H V2| = 4R2

(The symmedian point of a triangle is the point of concurrence of its three symmedian

lines The symmedian line through A is the reflection of the line through A and the cen-troid of A BC across the line bisecting angle B AC, and the symmedian lines through

B and C are defined similarly For more about the symmedian point, see Ross Hons-berger’s Episodes in Nineteenth and Twentieth Century Euclidean Geometry, MAA,

Washington, DC, 1995.)

Solution by Sin Hitotumatu, Kyoto University, Kyoto, Japan The points V satisfying

(1) lie on the Apollonius circles



V: |V A|

|V B| =

|BC|

|C A|

 and



V: |V A|

|V C| =

|BC|

|AB|



,

so there are at most two such points

The problem is not correct as stated We show that the two solution points that

satisfy (1) lie instead on the Euler line H O.

Let|BC| = a, |C A| = b, |AB| = c, −→ O A= a, −→O B= b, −→OC = c.

Lemma 1 If −→ O P = αa + βb + γ c, then −→ O P2 = R2(α + β + γ )2− (a2βγ +

b2γ α + c2αβ).

Proof With ·, · denoting the inner product,

−→O P2= (α2+ β2+ γ2)R2+ 2αβa, b + 2βγ b, c + 2γ αc, a.

Nowa, b = a b cos 2C = R2(1 − 2 sin2

C ) = R2− c2/2 by the law of sines;

similarly for the other two inner products Therefore,−→O P2 = R2(α2+ β2+ γ2+

2αβ + 2βγ + 2γ α) − c2αβ − b2γ α − a2βγ

Lemma 2 If ABC is an acute non-equilateral triangle, then 8R2 < a2+ b2+

c2 < 9R2.

Proof Since −→ O H = a + b + c, we conclude from Lemma 1 that |O H|2 = 9R2−

(a2+ b2+ c2) ≥ 0, with equality only if H = O (that is, ABC is equilateral) For the left inequality, writing s = (a + b + c)/2, we have

a2+ b2+ c2

(4Rs)2(a2+ b2+ c2)

= a2+ b2+ c2

a2b2c2 (−a4− b4− c4+ 2a2b2+ 2b2c2+ 2c2a2)

by Heron’s formula, and then

a2+ b2+ c2

R2 − 8 a2b2c2= (a2+ b2+ c2)− (a2+ b2+ c2)2

+ 4(a2+ b2+ c2)(a2b2+ b2c2+ c2a2) − 8a2b2c2

= (a2+ b2+ c2− 2a2)(a2+ b2+ c2− 2b2)(a2+ b2+ c2− 2c2),

which is positive since ABC is acute.

Now let −→O V = −→O H − λ(a + b + c), where λ is a parameter This is a parametric

description of the Euler line H O, since λ = 0 corresponds to V = H and λ = 1 cor-responds to V = O Since −→ AV = −→O V − −→O A = −λa + (1 − λ)b + (1 − λ)c, we see

by Lemma 1 that

|AV |2= R2(−λ + 1 − λ + 1 − λ)2− a2(1 − λ)2+ b2λ(1 − λ) + c2λ(1 − λ)

= R2(2 − 3λ)2+ λ(1 − λ)(a2+ b2+ c2) + (λ − 1)a2.

Therefore, if we can takeλ satisfying the equation

R2(2 − 3λ)2+ λ(1 − λ)(a2+ b2+ c2) = 0, (2) then we have |AV |2 = (λ − 1)a2 Similarly, for the same λ we have |BV |2 =

(λ − 1)b2and|CV |2= (λ − 1)c2, so we obtain the relation (1)

Let p (λ) be the quadratic expression on the left in (2) Then

p (λ) =9R2− (a2+ b2+ c2)λ2−12R2− (a2+ b2+ c2)λ + 4R2 The leading coefficient 9R2− (a2+ b2+ c2), call it ρ, is positive by Lemma 2 The

discriminant



12R2− (a2+ b2+ c2)2

− 16R2ρ

= (16 − 24)R2(a2+ b2+ c2) + (a2+ b2+ c2)2

= (a2+ b2+ c2)(a2+ b2+ c2− 8R2)

is positive by Lemma 2 This means that the equation (2) has two distinct rootsλ1and

λ2 Since p (0) > p(1) > 0, we have λ1, λ2 > 1 So λ1andλ2give the desired points

V1and V2, which lie on the Euler line H O Finally,

|H V1| |H V2| = λ1λ2a + b + c2= 4R2

ρ



9R2− (a2+ b2+ c2)= 4R2.

Also solved by GCHQ Problem Solving Group (U K.).

b T