THE AMERICAN MATHEMATICAL MONTHLY 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 February 28, 2009 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

11383 Proposed by Michael Nyblom, RMIT University, Melbourne, Australia Show

that

∞



n=1

cos−1



1+√n2+ 2n√n2+ 4n + 3

(n + 1)(n + 2)



= π

3.

11384 Proposed by Moubinool Omarjee, Lyc´ee Jean-Lurc¸at, Paris, France Let p n

denote the nth prime Show that

∞



n=1

(−1)√n

p n

converges

11385 Proposed by Jos´e Luis D´ıaz-Barrero, Universidad Polit´ecnica de Catalu˜na,

Barcelona, Spain Let α0,α1, andα2be the radian measures of the angles of an acute

triangle, and for i ≥ 3 let α i = α i−3 Show that

2



i=0

α2

i

α i+1α i+2



3+ 2 tan2α i

1/4 ≥ 3√3.

11386 Proposed by Greg Markowsky, Somerville, MA Consider a triangle A BC Let

O be the circumcircle of A BC, r the radius of the incircle, and s the semiperimeter.

Let arc(BC) be the arc of O opposite A, and define arc (C A) and arc(AB) similarly.

LetO A be the circle tangent to A B and AC and internally tangent to Oalong arc(BC),

and let R Abe its radius DefineO B,O C , R B , and R Csimilarly Show that

1

a R A

b R B

c R C

= s2

r abc

11387 Proposed by Oskar Maria Baksalary, Adam Mickiewicz University, Pozna´n,

Poland, and G¨otz Trenkler, Technische Universit¨at Dortmund, Dortmund, Germany.

Let C n ,n denote the set of n × n complex matrices Determine the shortest interval

[a, b] such that if P and Q in C n ,nare nonzero orthogonal projectors, that is, Hermitian

idempotent matrices, then all eigenvalues of PQ+ QP belong to [a, b].

11388 Proposed by M Farrokhi D.G., University of Tsukuba, Tsukuba Ibaraki, Japan.

Given a group G, let G2denote the set of all squares in G Show that for each natural number n there exists a finite group G such that the cardinality of G is n times the cardinality of G2

11389 Proposed by Elizabeth R Chen and Jeffrey C Lagarias, University of

Michi-gan, Ann Arbor, MI Given a multiset A = {a1, , a n } of n real numbers (not neces-sarily distinct), define the sumset S (A) of A to be {a i + a j : 1 ≤ i < j ≤ n}, a multi-set with n (n − 1)/2 not necessarily distinct elements For instance, if A = {1, 1, 2, 3},

then S (A) = {2, 3, 3, 4, 4, 5}.

(a) When n is a power of 2 with n ≥ 2, show that there are two distinct multisets A1

and A2such that S (A1) = S(A2).

(b) When n is a power of 2 with n ≥ 4, show that if r distinct multisets A1, , A r all

have the same sumset, then r ≤ n − 2.

(c∗) When n is a power of 2 with n ≥ 4, can there be as many as 3 distinct multisets with the same sumset?

(Distinct multisets are known to have distinct sumsets when n is not a power of 2.)

SOLUTIONS

Recalling 11159 and Sixty-Sixth Putnam A5

11277 [2007, 259] Proposed by Prithwijit De, University College Cork, Republic of

Ireland Find

 π/2

φ=0

 π/2

θ=0

log(2 − sin θ cos φ) sin θ

2− 2 sin θ cos φ + sin2θ cos2φ d θ dφ.

Solution by E Omey, EHSAL, Brussels, Belgium The answer is (π2/16) log 2 The

integral to be evaluated is

I =

 π/2

φ=0

 π/2

θ=0

f (sin θ cos φ) sin θ dθ dφ,

where f (x) = log(2 − x)/(2 − 2x + x2) In an editorial comment attached to the

so-lution of Problem 11159 in this MONTHLY114 (2007) 167, it was noted that if g is

integrable on[0, 1] then

 π/2

φ=0

 π/2

ψ=0

g (cos ψ cos φ) cos ψ dψ dφ = π

2

 1

0

g (t) dt.

After substitutingθ = π/2 − ψ and setting g = f , we see that I = (π/2)1

0 f (t)dt.

Write x = 1 − t The required computation,

 1 0

log(1 + x)

1+ x2 d x = π log 2

8 ,

appears in various tables and its evaluation was given as Problem A5 on the Sixty-Sixth William Lowell Putnam Competition (see this MONTHLY113 (2006) 733-743).

Also solved by R Bagby, D H Bailey & J M Borwein (U S A., Canada), D Beckwith, R Chapman (U K.),

H Chen, K Dale (Norway), B E Davis, G de Marco (Italy), A Fok (Hong Kong), O Furdui, M L Glasser,

J Grivaux (France), E A Herman, O Kouba (Syria), G Lamb, K McInturff, P Perfetti (Italy), O G, Ruehr, H.-J Seiffert (Germany), J G Simmonds, A Stadler (Switzerland), V Stakhovsky, R Stong, J Sun,

R Tauraso (Italy), GCHQ Problem Solving Group (U K.), Microsoft Research Problems Group, University

of Sharjah Problem Solving Group (United Arab Emirates), and the proposer.

Entire Limit

11278 [2007, 259] Proposed by Slavko Simic, Mathematical Institute SANU,

Bel-grade, Serbia Let f be a nonconstant entire function with nonnegative Taylor series

coefficients Prove that limr→∞ f (r)/r f(r) exists and is rational.

Solution by John H Lindsey II, Cambridge, MA First consider the case that f =

a0+ · · · + a n x n , a polynomial with a n > 0 We have

lim

f (r)

r f(r) = limr→∞

a0+ · · · + a n r n

a1r + · · · + a n nr n = lim

a0/r n + · · · + a n

a1/r n + · · · + a n n = 1

n

Now consider the case in which f = ∞k=0a k x kwith all coefficients nonnegative, and

not eventually zero For n ∈ N with a n > 0,

lim sup

f (r)

r f(r) ≤ lim supr→∞

n−1

k=0a k r k

a n nr n + lim sup

∞

k =n a k r k

∞

k =n a k kr k

≤ 0 + lim sup

∞

k =n a k r k

∞

k =n a k nr k = 1

n

Since n can be taken arbitrarily large, lim f (r)/r f(r) = 0.

Editorial comment It should be noted that the limit is intended in the sense of a real

variable r → +∞, not in the sense of a complex variable r → ∞.

Also solved by K F Andersen (Canada), J Bak, R Chapman (U K.), G De Marco (Italy), J Grivaux (France),

E A Herman, F Holland (Ireland), E J Ionascu, G Keselman, K.-W Lau (China), O P Lossers (Nether-lands), A Nakhash, N C Singer, R Stong, J Vinuesa (Spain), GCHQ Problem Solving Group (U K.), Mi-crosoft Research Problems Group, Northwestern University Math Problem Solving Group, and the proposer.

Sliding Beads

11279 [2007, 259] Proposed by Vitaly Stakhovsky, Redwood City, CA Two test-mass

beads are sliding along a vertical circular track under (Newtonian) constant gravity,

without friction The first bead M1 starts from the highest point on the circle, at some

nonzero velocity Some time later the second bead M2starts from the same position at

the top of the circle and with the same initial velocity as M1had Prove that there is a

circle to which the line through the current positions of M1and M2is always tangent, and find the center and radius of that circle in terms of the original circle

Solution by the proposer Assume that the original circle has unit radius and center

at the origin O Let OY point in the upward direction Describe the position of bead

i by its current location P i and the angle ϕ i = P i OY Assume both particles are

moving with ϕ i increasing Let v i = ˙ϕ i be the i th bead’s velocity The equation of

motion can be written asv2

i = 1 − μ cos ϕ ifor some 0≤ μ < 1 Note that v2

1− v2

2 =

μ(cos ϕ2− cos ϕ1) and ˙v i = (μ/2) sin ϕ i Hence

˙v1+ ˙v2

v2

1− v2 2

= 1

2· sinϕ1+ sin ϕ2

cosϕ2− cos ϕ1

= 1

2cot

ϕ1− ϕ2

2

.

Since|P1P2| = 2 sinϕ1 −ϕ2

2



we have

d

dt log|P1P2| = 1

2cot

ϕ1− ϕ2

2

(v1− v2) = ˙v1+ ˙v2

v1+ v2

= d

dt log(v1+ v2).

Thereforeτ = |P1P2|/(v1+ v2) is constant over time Note that

v2

1− v2

2 = μ sin ϕ1+ ϕ2

2

|P1P2|,

so we also haveμτ = (v1− v2)/ sin((ϕ1+ ϕ2)/2).

Let Q be the point on P1P2such that|P1Q |/|P2Q | = v1/v2 (Q is the instantaneous

center of rotation of the beads and will be the point of tangency of the circle.) Let a

perpendicular to P1P2 at Q intersect the vertical axis at a point D = (0, d) Let M

be the midpoint of P1P2 Now Q = (v1P2+ v2P1)/(v1+ v2) and M = (P1+ P2)/2,

hence

|M Q| = (v1− v2)|P1P2|

2(v1+ v2) =

v1− v2

2 τ = 1

2μτ2sin ϕ1+ ϕ2

2

.

Hence d = μτ2/2, and the point D is constant over time Also |DP i|2 = 1 + d2−

2d cos ϕ i = |DQ|2+ |Q P i|2, so

|DQ|2 = v21|DP2|2− v2

2|DP1|2

v2

1− v2 2

= 1 + d2− 2d v21cosϕ2− v2

2cosϕ1

v2

1− v2 2

= 1 + d2− 2

μ d

is constant over time Thus P1P2 is always tangent to the circle with center D and

radius

1+ d2− 2d/μ.

Note: In case P2starts when P1 is at the lowest point, we get r = 0 and hence the

points P1, P2, D are always collinear.

Also solved by ´ O Ciaurri & E Fern´andez & L Roncal (Brazil), J Freeman, J A Grzesik, J H Lindsey II,

J B Zacharias, and GCHQ Problem Solving Group (U K.).

What We Mean, Not What We Say

11280 [2007, 259] Proposed by Harris Kwong, SUNY Fredonia, Fredonia, NY Let f

be a positive nondecreasing function on the real line that is twice differentiable and

concave down For any list x of positive real numbers x1, , x n , let S= n

k=1x k In

terms of f and n, find

max

x n

k=1

( f (S − x k )) x k /S

Solution by GCHQ Problem Solving Group, Cheltenham, U K There is something

wrong with the problem statement: apart from constant functions, there are no positive concave functions on the real line, let alone twice differentiable nondecreasing ones

If n = 1 the answer is f (0) whatever properties f has For general n, we show that

if f is a positive nondecreasing function on the real line (not necessarily concave or

differentiable, or even continuous), then

sup

x n

k=1



f (S − x k )x k /S= lim

We claim first that the limit exists Indeed, if f is bounded above, its least upper bound

is its limit; if f is not bounded above, the limit is +∞ Let this limit be ϕ Next, if ϕ

is finite, thenn

k=1( f (S − x k )) x k /S≤n

k=1ϕ x k /S = ϕ Finally, if x1 = · · · = x n = z,

then

n

k=1



f (S − x k )x k /S = f ((n − 1)z)1/n = f ((n − 1)z) → ϕ

as z → ∞ This completes the proof of (1) The maximum is attained if and only if the limit is

There are of course functions on the nonnegative half of the real line with the full range of properties given in the problem statement The result given here applies to them

Editorial comment It seems that in the intended problem f is defined on the positive

real line, and n and S are given values Then maximize over all lists x satisfying the conditions Answer: f

(1 − 1/n)S

Also solved by R A Agnew, K F Andersen (Canada), J.-P Grivaux (France), E A Herman, J H Lindsey II,

V Stakhovsky, R Stong, M Tetiva (Romania), and Microsoft Research Problems Group.

A Triangular Result

11285 [2007, 358] Proposed by Yakub Aliyev, Qafqaz University and Baku State

Uni-versity, Baku, Azerbaijan Let six points be chosen in cyclic order on the sides of

triangle A BC: A1 and A2 on BC, B1and B2 on C A, and C1 and C2 on A B Let K denote the intersection of A1B2 and C1A2, L the intersection of B1C2 and A1B2, and

M the intersection of C1A2and B1C2 Let T , U , and V be the intersections of A1B2

and B1A2, B1C2 and B2C1, and C1A2 and C2A1, respectively Prove that lines AK ,

B L, and C M are concurrent if and only if points T , U , and V are collinear.

Solution I (one direction) by Apostolis Denis, Varvakeio High School, Athens, Greece.

We prove the first direction of the implication Suppose AK , B L, C M are concur-rent Write u v for the intersection of lines u and v (extended if necessary) Let X =

A1C2.B1A2, Y = B1A2.C1B2, and Z = C1B2.A1C2 Given triangles A BC and abc,

Desargues’ Theorem with its dual states that Aa , Bb, Cc are concurrent if and only

if A B ab, BC.bc, C A.ca are collinear We consider parallel lines to intersect at infinity

and allow for collinearity and concurrency at infinity (see Editorial comment below) Application of Desargues’ Theorem to the following three pairs of triangles gives the claimed implication:

(i) [ A BC and K L M]: AK , B L, C M are concurrent if and only if A B K L,

BC L M, C A.M K are collinear Since C1 lies on A B, and B2 lies on K L, we have

A B K L = C1B B2L Similarly, BC L M = B A2.B1L, and C A M K = B1B2.C1A2

(ii) [C1B A2and B2L B1]: C1B B2L, B A2.L B1, and A2C1.B1B2are collinear if and

only if C1B2, B L, A2B1are concurrent Note: Since C1B2.A2B1= Y , this concurrency

is equivalent to B L = LY Similarly, AK = K X and C M = M Z so K X, LY , M Z

are concurrent

(iii) [K L M and X Y Z ]: K X , LY , M Z are concurrent if and only if T = K L.XY ,

U = L M.Y Z, V = M K.Z X are collinear.

Solution II by The GCHQ Problem Solving Group, Cheltenham, United Kingdom

Pas-cal’s Theorem states that the intersection points of pairs of opposite sides of a hexagon

are collinear if and only if the vertices of the hexagon lie on a conic With the notation

of Solution I:

(a) Applying Desargues’ Theorem to A BC and K L M, we have: AK , BL, C M are

concurrent if and only if A B K L, BC.L M, C A.M K are collinear.

(b){AB.K L, BC.L M, C A.M K } = {C1C2.A1B2, A1A2.B1C2, B1B2.C1A2}, so by

Pascal’s Theorem these are collinear if and only if C1, C2, A1, A2, B1, B2 lie on a conic

(c){T, U, V } = {A1B2.B1A2, B1C2.B2C1, C1A2.C2A1}, so again by Pascal’s

The-orem, these are collinear if and only if A1, B2, B1, C2, C1, A2lie on a conic

Thus, AK , BL, C M are concurrent if and only if T, U, V are collinear.

Editorial comment The statement of the problem is true projectively, but for affine (or

Euclidean) geometry, it degenerates when X , Y , or Z fails to exist, or if K L M is a

degenerate triangle

Also solved by R Chapman (U K.), P D´alyay (Hungary), M Englefield (Australia), W Fosheng (China),

O P Lossers (Netherlands), M Tetiva (Romania), L Zhou, and the proposer.

Continuous Blackjack

11287 [2007, 359] Proposed by Stephen J Herschkorn, Highland Park, NJ Players

1 through n play “continuous blackjack.” At his turn, Player k considers a random number X kdrawn from the uniform distribution on[0, 1] He may either accept X kas

his score or draw a second number Y k from the same distribution, in which case his

score is X k + Y k if X k + Y k < 1 and 0 otherwise The highest score wins Give a rule

for when player k should draw a second number, in terms of k, n, the result of X k, and the highest score attained so far

Solution by BSI Problems Group, Bonn, Germany, and the editors We renumber the

players P0, P1, , P n−1 in reverse order, so that P k has k players yet to draw after his turn It turns out that the optimal strategy for P kis the same whether the remaining players play as a team or selfishly

Let Z k be P k ’s score; that is, Z k = X k if P k accepts his first number, Z k = X k + Y k

if he draws a second number Y k and X k + Y k < 1, and Z k = 0 if he draws a second

number and X k + Y k ≥ 1 For k = 0, , n − 1, let M k= maxj ≥k Z j , and let M n = 0

We assume selfish play That is, P k should take a second number if and only if

doing so improves his probability of winning, assuming P k−1through P0do the same

We ignore ties, which have zero probability Note that P k’s situation depends only on

the number k of players yet to draw and on M k+1 How M k+1was produced, and how

many players came earlier to produce it, are irrelevant

Let f k (z) denote the probability that if players P j for j ≥ k have already played and the best score so far is M k = z, then for all j < k, Z j < z—that is, all later players

achieve scores lower than z, so z is the winning score We will prove that for all k:

1 f kis a continuous, nondecreasing function on[0, 1] and f k (z) > 0 for z > 0.

2 There is a unique number g k ∈ [0, 1) such that f k (g k ) = 1

g k f k (t) dt, and P k

should take a second number if and only if X k < max(M k+1, g k ).

We first observe that statement 2 follows from statement 1 To see why, suppose statement 1 holds Clearly1

z f k (t) dt is a strictly decreasing continuous function of z,

0 f k (t) dt ≥ f k (0), and1

1 f k (t) dt = 0 < f k (1), so by the intermediate value

theo-rem there is a unique number g k ∈ [0, 1) such that f k (g k ) =1

g k f k (t) dt Furthermore,

if z < g k then1

z f k (t) dt > f k (z), and if z > g k then1

z f k (t) dt < f k (z) To

deter-mine how player P k should play, suppose that all players P j for j > k have already

played Clearly if X k < M k+1then player P k must take a second number to have any

hope of winning If X k > M k+1, then player P k ’s probability of winning is f k (X k ) if

he doesn’t take a second number, and1

X k f k (t) dt if he does The latter probability is

larger than the former if and only if X k < g k , so P k should take a second number if

and only if X k < max(M k+1, g k ).

To prove statement 1, we proceed by induction on k In the case k= 0, if all of the

players have played and M0= z, then z is definitely the winning score, so f0(z) = 1.

Now assume statement 1 (and therefore also statement 2) holds for k Suppose that players P j for j ≥ k + 1 have already played and M k+1 = z There are two ways for player P k to end up with a score Z k < z: either X k + Y k < z, or X k + Y k > 1 The

probability of the first event isz

0(z − x) dx = z2/2, and the probability of the second

ismax(z,g k )

0 x d x = (max(z, g k ))2/2 Thus, the probability that Z k < z is:

h k (z) =



1

2(g2

k + z2), if z ≤ g k;

z2, if z > g k

If Z k < z then M k = z, and therefore the probability that Z j < z for all j < k is

f k (z) Thus, we have f k+1(z) = h k (z) f k (z), and combining our formula for h k (z) with

statement 1 for k, we see that statement 1 holds for k+ 1 as well

Finally, we wish to determine the numbers g k We first observe that for all k, g k+1 >

g k To see why, suppose that g k+1 ≤ g k Then, on putting l = k + 1, we have

f l (g l ) = h k (g l ) f k (g l ) ≤ h k (g l ) f k (g k ) = h k (g l )

 1

g k

f k (t) dt

≤ h k (g l )

 1

g l

f k (t) dt <

 1

g l

h k (t) f k (t) dt =

 1

g l

f l (t) dt = f l (g l ),

a contradiction It follows that g k+1 > g k, as claimed

Next, we observe that for z ≥ g k , f k (z) = z 2k This follows easily by induction

from the facts that f0(z) = 1, f k+1(z) = h k (z) f k (z), and g k+1 > g k, and the formula

for h k (z).

Finally, we use this formula for f k (z) to get an equation that we can solve to find

g k:

g k 2k = f k (g k ) =

 1

g k

f (t) dt =

 1

g k

t 2k dt = 1

2k+ 1(1 − g

2k+1

In summary, the best strategy is for P k to take a second number if and only if

X k < max(M k+1, g k ), where g k is the unique nonnegative solution to z 2k = (1 −

z 2k+1)/(2k + 1).

Editorial comment If the remaining players are not selfish, but cooperate to try to stop

P k , this is still P k ’s best strategy For j < k, player P j now accepts any score X j that

is better than M j The result is that h j (z) = z2 and f j (z) = z 2 j for j < k Thus P k

should still take a second number if and only if X k < max(M k+1, g k ).

Also solved by D R Bridges, J Chachulska & W Matysiak (Poland), R Chapman (U K.), J Freeman, J H Lindsey II, G Pastor (Mexico), T Rucker, B Schmuland (Canada), R Staum, T Tam, E A Weinstein, GCHQ Problem Solving Group (U K.), and the proposer.

Formulas Involving the Angles of a Triangle

11289 [2007, 359] Proposed by Oleh Faynshteyn, Leipzig, Germany Let A BC be a

triangle with sides a, b, and c, all different, and corresponding angles α, β, and γ

Show that

(a)(a + b) cot(β +1

2γ ) + (b + c) cot(γ +1

2α) + (a + c) cot(α +1

2β) = 0.

(b) (a − b) tan(β + 1

2γ ) + (b − c) tan(γ + 1

2α) + (c − a) tan(α +1

2β) = 4(R + r),

where R is the circumradius of the triangle and r the inradius.

Solution by Marian Tetiva, Birlad, Romania Since α + β + γ = π, we have β +

1

2γ = π

2 − α−β

2 , so cot(β +1

2γ ) = tan α−β

2 and tan(β +1

2γ ) = cot α−β

2 From the law

of sines and trigonometric identities, we get

a + b = 2R(sin α + sin β) = 4R sin α + β

2 cos

α − β

2 ,

a − b = 2R(sin α − sin β) = 4R sin α − β

2 cos

α + β

2 .

Similar equations are obtained by cyclically permuting a , b, c and α, β, γ

We will write “

” for a sum with three terms obtained by cyclic permutation from

the one term given For part (a):



(a + b) cot β +1

2γ

= 4Rsinα + β

2 cos

α − β

2 tan

α − β

2

= 4Rsinα + β

2 sin

α − β

2 = 4R sin2 α

2 − sin2β

2

= 0.

For part (b):



(a − b) tan β +1

2γ

= 4Rsinα − β

2 cos

α + β

2 cot

α − β

2

= 4Rcosα + β

2 cos

α − β

2 = 4Rcosα = 4(R + r).

For the last step we used the formula

 cosα = 1 + 4 sin α

2sin

β

2sin

γ

2 = 1 + r

R

Also solved by Z Ahmed (India), S Amghibech (Canada), M Bataille (France), R Chapman (U.K.), P P D´alyay (Hungary), P De (Ireland), A Demis (Greece), A Fok (China), W Fosheng (China), V V Garc´ıa (Spain), M Goldenberg & M Kaplan, O Kouba (Syria), H Kwong, K.-W Lau (China), O P Lossers (Nether-lands), Y Mikata, E Mouroukos (Greece), R D Nelson (U.K.), J H Nieto (Venezuela), P E Nuesch (Switzer-land), C R Pranesachar (India), J Rooin & A Alikhani (Iran), V Schindler (Germany), L Sega, H.-J Seiffert (Germany), M Shattuck, R A Simon (Chile), A Stadler (Switzerland), D Vacaru (Romania), M Vowe (Switzerland), L Zhou, GCHQ Problem Solving Group (U.K.), Microsoft Research Problems Group, North-western University Math Problem Solving Group, and the proposer.

There are of course functions on the nonnegative half of the real line with the full range of properties given in the. .. statement The result given here applies to them

Editorial comment It seems that in the intended problem f is defined on the positive

real line, and n and S are given values Then... Solving Group, Cheltenham, U K There is something

wrong with the problem statement: apart from constant functions, there are no positive concave functions on the real line, let alone twice