SOLUTIONS - cms.math.ca · 270/ Solutions SOLUTIONS No problem is ever permanently closed. The editor is always pleased to consider for publication new solutions or new insights on

270/ Solutions

SOLUTIONSNo problem is ever permanently closed. The editor is always pleased to consider forpublication new solutions or new insights on past problems.

Statements of the problems in this section originally appear in 2018: 44(6), p. 256–259;44(7), p. 302–305; 44(8), p. 340–343.

4351. Proposed by Miguel Ochoa Sanchez and Leonard Giugiuc.

Let ABC be a triangle and let ω be its incircle. Let E and F be the tangency pointsof ω and the sides AC and AB, respectively. Let G be the second intersection pointof ω and BE and, similarly, let D be the second intersection point of ω and CF .Prove that

FE ·GDFG · ED = 3.

We received 6 submissions, all of which were correct, and will feature two of them.

Solution 1, by Richard B. Eden.

Let I be the incenter of 4ABC and K the point on FD such that ∠KGF =∠DGE. Since ∠GFK = ∠GED, then

4KGF ∼ 4DGE.Next, since ∠EGF = ∠EGK + ∠KGE = ∠EGK + ∠DGE = ∠DGK and∠FEG = ∠KDG, then

4KDG ∼ 4FEG.From the similarity of these triangles, we get FK

FG = EDEG and DK

DG = EFEG , so

FE ·GDFG · ED =

DK · EGFK · EG =

DK

FK.

Crux Mathematicorum, Vol. 45(5), May 2019

Solutions /271

Our goal, therefore, is to prove that DKFK = 3. Let S be the midpoint of DF so

that the problem is reduced to showing that K is the midpoint of FS. Let T bethe midpoint of EG, and M the point of contact of ω and BC. First, we will showthat 4TGF ∼ 4EMF . We have ∠TGF = ∠EGF = ∠EMF . Since T is themidpoint of chord EG of ω,

∠ITB = ∠ITG = 90◦ = ∠IFB = ∠IMB,

so F , B, M and T are concyclic. Therefore,

∠FTG = ∠FTB = ∠FMB = ∠FEM.

Therefore, 4TGF ∼ 4EMF . Similarly, 4EDS ∼ 4EMF . Therefore, we have4TGF ∼ 4EDS.

It follows thatTG

GF=ED

DS, so

ET

FG=ED

FS. Since

∠GFS = ∠GFD = ∠GED = ∠TED,

it follows that 4GFS ∼ 4TED. Now let J be the midpoint of FG. Since4GFK ∼ 4GED and the medians KJ and DT correspond to each other, then4JFK ∼ 4TED. This implies 4JFK ∼ 4GFS, with vertices written in corre-sponding order. Since J is the midpoint of FG, K is the midpoint of FS.

Solution 2, by Michel Bataille.

We shall use complex numbers, denoting by m (small letter) the complex affix ofthe point M (capital letter). Without loss of generality, we suppose that ω is theunit circle. We observe that if M and N are points of ω, then

MN2 = |m− n|2 = (m− n)(m− n) = (m− n)

(1

m− 1

n

)= − (m− n)2

mn.

It readily follows that

FE2 ·GD2

FG2 · ED2=

(e− f)2(d− g)2

(g − f)2(d− e)2

and therefore our problem amounts to showing that the ratio ρ =(e− f)(d− g)

(g − f)(d− e)equals 3 or −3.

Let W be the point of tangency of ω and BC. From the respective equationsz+w2z = 2w and z+e2z = 2e of the lines BC and CA (which are the tangents to

ω at W and E), we obtain c =2ew

e+ w. Since C is on the line DF whose equation

is z + dfz = d+ f , we have c+ dfc = d+ f and a short calculation yields

d =2we− wf − efe+ w − 2f

=w(e− f) + e(w − f)

(e− f) + (w − f)=w(e− f) + e(w − f)

α

Copyright © Canadian Mathematical Society, 2019

272/ Solutions

(with α = (e− f) + (w − f)). Exchanging e and f gives

g =w(f − e) + f(w − e)

(f − e) + (w − e) =w(f − e) + f(w − e)

β

(with β = (f − e) + (w − e)). Then we easily calculate

g − f = (β)−1(f − e)(w − f),

d− e = (α)−1(e− f)(w − e), and

d− g = (αβ)−1(e− f)(3w2 − 3we− 3wf + 3ef).

This leads to

ρ =3(e− f)2(w2 − we− wf + ef)

(f − e)(w − f)(e− f)(w − e) = −3,

and the result follows.

4352. Proposed by Thanos Kalogerakis.

Explain how to locate six points A,B,C,D,E, F in that order about the circum-ference of a circle so that the resulting convex hexagon has an incircle, yet is notregular.

The solutions from C.R. Pranesachar and the proposer were the only submissionswe received. We present a composite of their work.

A polygon that has both a circumcircle and an incircle is called bicentric. Itis clear that for a convex hexagon to have an incircle, it is necessary that theangle bisectors at all six vertices be concurrent. (The point of concurrence willbe equidistant from all six sides.) Our construction of a bicentric hexagon will bebased on the following theorem, which is interesting in its own right.

Theorem. If ABCD is a cyclic quadrilateral, then the bisectors of the angles at Band at C meet at a point of the chord AD if and only if there exists a point S ofAD for which AS = AB and DS = DC.

First, assume that for a point S of AD we have DS = DC. Assume further(without loss of generality) that the points have been labeled so that AB ≤ CD.


Solutions /273

Define P to be the second point where the circle BCS intersects the line AD. Weshow that BP bisects ∠ABC.

From BSPC cyclic, we have ∠PBC = ∠PSC.

From ∆DSC isosceles, we have ∠PSC = ∠DSC = ∠DCS =180◦ − ∠SDC

2.

From ABCD cyclic, we have ∠ABC = 180◦ − ∠SDC.

It follows that ∠PBC = 12∠ABC; that is, BP bisects ∠ABC, as claimed. Simi-

larly, CP bisects ∠BCD.

For the converse, we assume that the bisectors of the angles at B and at C intersectat a point P of the chord AD and, moreover, that S is the point of AD for whichAS = AB. We will show that DS = DC. We have proved above that thequadrilateral BSPC is cyclic. Similarly, for the point S′ for which DC = DS′ wehave BS′PC is also cyclic. But there is just one point of AD other than P thatcan lie on the circle BCP , whence we conclude that S′ = S, which finishes theproof.

Finally, for an example of a bicentric hexagon ABCDEF that is not regular,denote its circumcircle by γ, let AD be a diameter, and choose a point S of ADdifferent from its midpoint. Define B and F to be the points where γ intersectsthe circle (A,AS) (with center A and radius AS); define C and E to be the pointswhere γ intersects the circle (D,DS) labeled so that B and C are on the sameside of AD. By our theorem, the bisectors of the angles at B and C meet in apoint of AD as do the bisectors of the angles at E and F . By symmetry all fourof those angle bisectors must meet in the same point of AD while AD bisects theangles at both endpoints. In other words, all six angle bisectors are concurrent inthe center of the incircle of ABCDEF . This concludes the required construction.

Further comments. Once we have constructed one bicentric hexagon, we can con-struct infinitely many of them according to the Great Poncelet Theorem for Circles:If there is an n-sided polygon inscribed in a circle α and circumscribed about acircle β, then for any point A of α there exists an n-sided polygon, also inscribedin α and circumscribed about β, which has A as one of its vertices. In fact, thetheorem applies more generally to families of conics, but the proof, even in thesimplest case of a pair of bicentric polygons, is not easy. Note that Poncelet’stheorem combined with our result provides a recipe for the construction of all bi-centric hexagons — given any bicentric hexagon with circumcircle α and incircleβ, there exists a bicentric hexagon ABCDEF inscribed in α and circumscribedabout β having AD a diameter of α. Simply start the Poncelet construction withA on the line containing the two centers.

Editor’s comments. For the “simplest” proof of Poncelet’s theorem known to thiseditor, see the article “A Simple Proof of Poncelet’s Theorem” by Lorenz Halbeisenand Norbert Hungerbuhler, The American Mathematical Monthly 122:6 (January2014) 537-551.


274/ Solutions

4353. Proposed by Michel Bataille.

Evaluate

limn→∞

1

n

∞∑k=1

n∑j=1

1

k(j+k−1j

) .We received four solutions. We present the one by Bao Do, lightly edited.

Let

s(j) =

∞∑k=1

1

k(j+k−1j

) =

∞∑k=1

j!

(k + j − 1) · · · (k + 1)k2.

Note that the sum in the definition converges, since

∞∑k=1

1

k(j+k−1j

) ≤ ∞∑k=1

1

k2=π2

6.

We need to evaluate limn→∞1n

∑nj=1 s(j). Consider

s(j)

j=

∞∑k=1

(j − 1)!

(k + j − 1) · · · (k + 1)k2

= (j − 2)!

∞∑k=1

(1

(k + j − 2) · · · (k + 1)k2− 1

(k + j − 1) · · · (k + 1)k

)

=1

j − 1

∞∑k=1

(j − 1)!

(k + j − 2) · · · (k + 1)k2− (j − 2)!

∞∑k=1

1

(k + j − 1) · · · k

=s(j − 1)

j − 1− (j − 2)!

(j − 1)

∞∑k=1

(1

(k + j − 2) · · · k −1

(k + j − 1) · · · (k + 1)

)=s(j − 1)

j − 1− 1

(j − 1)2.

We rewrite this ass(j)

j− s(j − 1)

j − 1= − 1

(j − 1)2.

We take the sum from j = 2 to n on both sides to obtain

n∑j=2

(s(j)

j− s(j − 1)

j − 1

)= −

n∑j=2

1

(j − 1)2

=⇒ s(n)

n− s(1)

1= −

n−1∑j=1

1

j2

=⇒ s(n) = n(s(1)−

n−1∑j=1

1

j2

)= n

(π2

6−n−1∑j=1

1

j2

).


Solutions /275

Finally, we apply the Stolz-Cesaro theorem twice to calculate the limit [Ed.: Theeditor found that most statements of Stolz-Cesaro require the sequence used forthe denominators to be strictly monotone and divergent. However, the case whenboth denominator and numerator sequences converge to 0 and the denominatorsequence is strictly monotone holds as well.]:

limn→∞

1

n

n∑j=1

s(j) = limn→∞

∑n+1j=1 s(j)−

∑nj=1 s(j)

(n+ 1)− n = limn→∞

s(n+ 1)

= limn→∞

(n+ 1)

(π2

6−

n∑j=1

1

j2

)= limn→∞

π2

6 −∑nj=1

1j2

1n+1

= limn→∞

−∑n+1j=1

1j2 +

∑nj=1

1j2

1n+1 − 1

n

= limn→∞

− 1(n+1)2

− 1n(n+1)

= limn→∞

n(n+ 1)

(n+ 1)2= 1.

4354. Proposed by Ruben Dario Auqui and Leonard Giugiuc.

Let ABCD be a square with side length 1. Consider points M ∈ AB, N ∈ BCand P ∈ CA such that the triangles BMN and PMN are congruent. Prove that

1

MB+

1

BN= 2 +

2

MB +BN.

Thirteen correct solutions were received, along with one incorrect one. We providea sample of the variety of approaches used.

Solution 1.

The triangles are congruent through a reflection in the axis MN . Assign coordi-nates with A ∼ (0, 1), B ∼ (0, 0), C ∼ (1, 0), M ∼ (0, a), N ∼ (c, 0). The line MNhas equation (x/c) + (y/a) = 1 and slope −a/c. The line BP is perpendicular toMN and has slope c/a. Hence P ∼ (a(a+ c)−1, c(a+ c)−1). Since BP and MNintersect at the midpoint of BP , we have

a

2c(a+ c)+

c

2a(a+ c)= 1

⇐⇒ a2 + c2 = 2ac(a+ c)

⇐⇒ (a+ c)2 = 2ac(a+ c+ 1)

⇐⇒ 1

a+

1

c=a+ c

ac= 2

(1 +

1

a+ c

)as desired.


276/ Solutions

Solution 2, by Ivko Dimitric.

Using the notation and initial results from Solution 1, along with the formula forthe distance of a point to the line MN with equation ax + cy − ac = 0, we findthat

dist(B,MN) =ac√a2 + c2

and

dist(P,MN) =|a2(a+ c)−1 + c2(a+ c)−1 − ac|√

a2 + c2=

1√a2 + c2

(a2 + c2

a+ c− ac

),

since a2 + c2−ac(a+ c) = a2(1− c)+ c2(1−a) > 0. (The first distance can also befound by taking the area of triangle MBN in two ways.) Since the two distancesare equal,

a2 + c2

a+ c= 2ac⇔ (a+ c)2 = 2ac(a+ c) + 2ac,

from which the result follows.

Solution 3, by I.J.L. Garces.

Let a and c be the respective lengths of BM and BN , and let ∠BMN = θ, sothat c = a tan θ.

1

a+

1

c− 2

a+ c=

1

a

(1 +

cos θ

sin θ− 2 cos θ

sin θ + cos θ

)=

1

a sin θ(cos θ + sin θ).

Observe that MP = MB and ∠APM = 2θ − 45◦. Use the Sine Law on triangleAMP to obtain

1

a√

2=

sin(2θ − 45◦)

1− a ,

whence1

a= sin 2θ − cos 2θ + 1 = 2 sin θ(cos θ + sin θ).

Hence1

a+

1

c− 2

a+ c= 2,

as desired.

Solution 4, by Cristobal Sanchez-Rubio.

Let Q be the foot of the perpendicular from P to AB, and let a, c, d be therespective lenghts of BM , BN , PQ. The altitudes of triangles PAM and PNChave respective lengths d and 1− d. With [. . . ] denoting area, we have that

1

2= [ABC] = [PAM ] + [PMBN ] + [PNC]

= [PAM ] + 2[MBN ] + [PNC]

=1

2(1− a)d+ ac+

1

2(1− c)(1− d)


Solutions /277

Hence d(a− c) = c(2a− 1).

Since QP‖BN and BP ⊥MN ,

∠QPB = ∠PBN = 90◦ − ∠MBP = ∠BMN.

It follows that the right triangles PQB and MBN are similar, so d/(1− d) = a/cand d = a/(a+ c). Plugging this into the previous equation yields that a(a− c) =c(a+ c)(2a− 1) which can be unravelled to the desired result

1

a+

1

c= 2 +

2

a+ c.

Editor’s comments. For the situation to be viable, P must be distant no morethan 1 from each of the vertices A and C. If the triangles BMN and PMNare congruent with each of M and N corresponding to the other, then BNPMis a rectangle and MB + BN = 1. In this case, the equation holds only whenMB = BN and P is the midpoint of AC.

4355. Proposed by Mihaela Berindeanu.

Let H be the orthocenter of triangle ABC with P the midpoint of AB and Q themidpoint of AC. If WY is the line perpendicular to HP at H with W ∈ AC andY ∈ BC, while XZ is the perpendicular to HQ at H with X ∈ AB and Z ∈ BC,prove that the quadrilateral WXY Z is a parallelogram.

In rewording the proposer’s problem, the editors mistakenly omitted the necessarycondition that WY and XZ both had to pass through H. All four submissionscorrected that error. We present the solution by Leonard Giugiuc.

To prove that WXY Z is a parallelogram it is sufficient to prove that H is themidpoint of both diagonals. To that end, we introduce Cartesian coordinates,choosing (without loss of generality)

A(0, 1), B(−u, 0), and C(v, 0),


278/ Solutions

where u = cotB and v = cotC. We therefore have

H(0, uv), P

(−u

2,

1

2

), and Q

(v

2,

1

2

).

The slope of the line PH is 2uv−1u , so the equation of WY is (1−2uv)(y−uv) = ux.

Also, BC is y = 0 and AC is x+ vy = v. Thus, from

(1− 2uv)(y − uv) = ux and x+ vy = v

we get W (v(1− 2uv), 2uv), while from

(1− 2uv)(y − uv) = ux and y = 0

we get Y (v(2uv − 1), 0). We deduce that H is the midpoint of the segment WY .Similarly (by interchanging the roles of −u and v), we find X (u(2uv − 1), 2uv)and Z (u(1− 2uv), 0), so that H is the midpoint of XZ, which concludes the proof.

4356. Proposed by Leonard Giugiuc and Diana Trailescu.

Solve the following system over reals:a+ b+ c+ d = 6,

a2 + b2 + c2 + d2 = 12,

abc+ abd+ acd+ bcd = 8 + abcd.

We received 14 submissions including that from the proposers. All are correct andwe present the solution by Ramanujan Srihari, modified slightly by the editor.

From the first two equations we readily obtain

ab+ ac+ ad+ bc+ bd+ cd = 12.

Let abcd = k and let

f(x) = x4 − 6x3 + 12x2 − (k + 8)x+ k (1)

be the polynomial with a, b, c, d as its roots.

Thenf ′(x) = 4x3 − 18x2 + 24x− (k + 8). (2)

Suppose f ′(r) = 0 where r ∈ R. Then the slope of y = f(x) at x = r is 0.

Now, by straightforward computations we have

f(r) = f(r) + (1− r)f ′(r)= r4 − 6r3 + 12r2 − (6 + 8)r + k + (1− r)(4r3 − 18r2 + 24r − (k + 8))

= −3r4 + 16r3 − 30r2 + 24r − 8

= −(r2 − 4r + 4)(3r2 − 4r + 2)

= −(r − 2)2(3r2 − 4r + 2).


Solutions /279

Since 3r2− 4r+ 2 = 3((r− 23 )2 + 2

9 ) > 0 we see that f(r) ≤ 0 with f(r) = 0 if andonly if r = 2.

Assume that f ′(2) 6= 0. Then for all t with f ′(t) = 0 we have f(t) < 0. Thisimplies that f has only two simple roots which is a contradiction since f(x) hasfour roots, a, b, c, and d.

Thus, f ′(2) = 0 and t = 2 is the only multiple root of f . From (2), f ′(2) = 0 =⇒k = 0.

Hence we get from (1) that f(x) = x4 − 6x3 + 12x2 − 8x = x(x− 2)3 and so

(0, 2, 2, 2), (2, 0, 2, 2), (2, 2, 0, 2), and (2, 2, 2, 0)

are the only solutions.

4357. Proposed by Arsalan Wares.

Suppose ABCD represents a 9 by 12 rectangular sheet. Points K and L aremidpoints of sides AB and DC, respectively. First, edge AD, of the rectangularsheet ABCD, is folded over by making a crease along DK. Then edge BC isfolded over by making a crease along BL. Folded corners of the sheet overlap overa polygonal region C ′XA′Y as shown.

Find the area of the overlapping polygon C ′XA′Y .

We received 14 correct solutions and 4 incorrect submissions. We present thesolution by Skidmore College Problem Group.

We claim that the area of C ′XA′Y is2160

169square units.

Note first that∠CBL = ∠C ′BL = ∠ADK = ∠KDA′.

Thus∠ABL = 90◦ − ∠CBL = ∠AKD = 90◦ − ∠ADK.

But ∠ABL = ∠AKD, implying that DK and BL are parallel. Thus, by alternat-ing interior angles, ∠KC ′B = ∠C ′BL. Now

∠C ′XK = 180◦ − ∠KCB′ − ∠C ′KX = 90◦

and ∠XC ′L = ∠KA′D = 90◦, so C ′B and DA′ are parallel, implying thatC ′XA′Y is a rectangle.


280/ Solutions

We know that AB = DC = 12, BC = AD = 9, and CL = 6. Set a = A′Y = C ′Xand b = XA′ = C ′Y . Then DY 2 + Y L2 = DL2, so

(9− a)2 + (6− b)2 = 36,

i.e.,a2 − 18a+ b2 − 12b+ 81 = 0.

But ∆C ′DY is similar to ∆KDA′, so

9− a9

=b

6,

giving a = 9− 32b. Thus(

9− 3

2b

)2

− 18

(9− 3

2b

)+ b2 − 12b+ 81 = 0,

which, after simplifying, is13

4b2−12b = 0, implying that b =

48

13and thus a =

45

13.

This gives the area of2160

169.

4358. Proposed by George Stoica.

Let f : [0,∞)→ [0,∞) be a non-increasing function such that∫ ∞0

f(x) sin(2πx)dx = 0.

Prove that f is constant on each of the intervals (n, n+ 1), n ∈ N.

We received 13 submissions, including the one from the proposer. Among them 12are correct. The other solver apparently misread the question and assumed thatthe function is non-decreasing instead, but arrived at the same conclusion! Wepresent the solution by Kee-Wai Lau.

For any k ∈ N, let ak =∫ k0f(x) sin(2πx)dx. Then

ak =

k−1∑n=0

(∫ n+1/2

n

f(x) sin(2πx)dx+

∫ n+1

n+1/2

f(x) sin(2πx)dx

)

=

k−1∑n=0

(Jn +Kn) =

k−1∑n=0

In,

where In = Jn+Kn, Jn =∫ n+1/2

nf(x) sin(2πx)dx, andKn =

∫ n+1

n+1/2f(x) sin(2πx)dx.

Substituting x = n+ y and x = n+ y+ 1/2 into Jn and Kn, respectively, we thenobtain

In =

∫ 1/2

0

(f(n+ y)− f(n+ y + 1/2)) sin(2πy)dy. (1)


Solutions /281

Since f is non-increasing, we have In ≥ 0 and since limk→∞ ak = 0, we obtainIn = 0 for any n ∈ N ∪ {0}. It follows from (1) that except for a set of measurezero,

f(n+ y)− f(n+ y + 1/2) = 0 for y ∈ (0, 1/2).

Let ε ∈ (0, 1/2). Then there exist

t0 ∈ (n+ 1/2, n+ 1/2 + ε) and t1 ∈ (n+ 1− ε, n+ 1)

such that f(t0) = f(t0 − 1/2) and f(t1) = f(t1 − 1/2). Since f is non-increasing,

f(y) = f(n+ 1/2) for y ∈ [n+ ε, n+ 1/2] ∪ [n+ 1/2, n+ 1− ε]= [n+ ε, n+ 1− ε].

Letting ε → 0+ we then obtain f(y) = f(n + 1/2) for all y ∈ (n, n + 1). Thus, fis constant on each of the intervals (n, n+ 1), n ∈ N.

4359. Proposed by Daniel Sitaru.

Let a, b and c be positive real numbers. Prove that

3 ln(ab + bc + ca) +a

c+b

a+c

b≥ a+ b+ c+ ln 27.

We received 10 submissions, including the one from the proposer, all of which arecorrect. We present a composite based on the solutions given by Richard B. Edenand Ramanujan Srihari, which are similar to all the other solutions submitted.

Let f(x) = lnx, x > 0. Then f ′′(x) = − 1

x2< 0 so f is concave. By Jensen’s

Inequality we then have

3 ln(ab + bc + ca

3

)≥ b ln a+ c ln b+ a ln c (1)

with equality if and only if a = b = c.

Next, consider g(x) = lnx + 1x − 1, x > 0. Then g′(x) =

x− 1

x2which implies

g(x) ≥ g(1) = 0 so lnx ≥ 1− 1x for all x > 0. Hence,

b ln a ≥ b(1− 1a ), c ln b ≥ c(1− 1

b ), a ln c ≥ a(1− 1c ). (2)

From (1) and (2), we then obtain

3 ln(ab + bc + ca) ≥ b(1− 1a ) + c(1− 1

b ) + a(1− 1c ) + ln 27

so

3 ln(ab + bc + ca) +a

c+b

a+c

b≥ a+ b+ c+ ln 27

follows, completing the proof.


282/ Solutions

4360. Proposed by H. A. ShahAli.

Let a, b, c be non-negative real numbers such that a+b+c = 1. Find the minimumand maximum values of the expression

a+ b

1 + ab+

b+ c

1 + bc+

a+ c

1 + ac.

When do those extreme values occur?

We received 9 correct solutions and 3 incorrect submissions. We present the solu-tion by Paolo Perfetti, modified by the editor.

We note first that

a+ b

1 + ab+

b+ c

1 + bc+

a+ c

1 + ac≤ a+ b+ b+ c+ c+ a = 2,

with equality for a = 1, b = c = 0 and its permutations, so the maximum is 2.

The minimum is 9/5 and it attained for a = b = c = 1/3 or a = b = 1/2, c = 0and its permutations. To prove this we show that

a+ b

1 + ab+

b+ c

1 + bc+

a+ c

1 + ac≥[a+ b

1 + ab+

b+ c

1 + bc+

a+ c

1 + ac

] ∣∣∣a+b2 , a+b

2 ,c=0.

Equivalently, we need to prove that

a+ b

1 + ab+

b+ c

1 + bc+

a+ c

1 + ac≥ a+ b

1 + (a+b)2

4

+a+ b

2+a+ b

2=

a+ b

1 + (a+b)2

4

+ 1.

To do so, it suffices to show that

a+ b

1 + ab≥ a+ b

1 + (a+b)2

4

andb+ c

1 + bc+

a+ c

1 + ac≥ 1.

The first of these follows from the AM-GM inequality since ab ≤ (a+b)2

4 . Thesecond can be successively rewritten as

(b+ c)(1 + ac) + (a+ c)(1 + bc) ≥ (1 + bc)(1 + ac)

(a+ b) + 2c+ 2abc+ (a+ b)c2 − 1− c(a+ b)− abc2 ≥ 0

1 + c+ 2abc+ (a+ b)c2 − 1− c(a+ b)− abc2 ≥ 0

c[1 + ab(2− c) + c(1− c)− (1− c)] ≥ 0

c[ab(2− c) + c(2− c)] ≥ 0

c(2− c)(ab+ c) ≥ 0,

which is true. It follows that

a+ b

1 + ab+

b+ c

1 + bc+

a+ c

1 + ac≥[a+ b

1 + ab+

b+ c

1 + bc+

a+ c

1 + ac

] ∣∣∣12 ,

12 ,0

=9

5.


Solutions /283

4361. Proposed by Andrew Wu.

Let ABC be a scalene triangle with circumcircle Γ, circumcenter O and incenter I.Suppose that L is the midpoint of the arc BAC of Γ. The perpendicular bisector ofAI meets at X the arc AC that contains B, and at Y the arc AB that contains C.Let XL and AC meet at P ; let Y L and AB meet at Q. Show that the orthocenterof triangle OPQ lies on XY .

Three of the four submissions that we received were correct, and one was flawed.We feature the solution by Andrea Fanchini.

We use barycentric coordinates with respect to triangle ABC, where its circum-circle Γ is described by the equation a2yz+ b2zx+ c2xy = 0. It will turn out thatthe orthocenter of ∆OPQ, call it H ′, is the intersection of XY and BC.

The midpoint L of the arc BAC of Γ is

L = OMa ∩ Γ =(a2 : −b(b− c) : c(b− c)

).

The perpendicular bisector of AI is

XY : bcx− c(a+ c)y − b(a+ b)z = 0,

whence the points X and Y are

X = XY ∩ Γ =(a(a+ b) : b(a+ b) : −c2

),

Y = XY ∩ Γ =(a(a+ c) : −b2 : c(a+ c)

).

Therefore, the lines XL and Y L are

XL : c(c− b)x+ acy + a(a+ b)z = 0,

Y L : b(b− c)x+ a(a+ c)y + abz = 0,


284/ Solutions

so that the points P and Q are

P = XL ∩AC = (a(a+ b) : 0 : c(b− c)) ,Q = Y L ∩AB = (a(a+ c) : b(c− b) : 0) .

Finally, two altitudes of ∆OPQ are given by the polars with respect to Γ of Qand of P , namely

POQ∞⊥ : bc(b− c)x− ac(a+ c)y − ab(a+ b)z = 0, and

QOP∞⊥ : bc(b− c)x+ ac(a+ c)y + ab(a+ b)z = 0.

Their intersection is the orthocenter of triangle OPQ, namely

H ′ (0 : −b(a+ b) : c(a+ c)) .

One easily checks that it lies on XY (obtained above), as desired. As a bonus, wesee that H ′ also lies on the line BC.

Editor’s comments. Note the relationship between our Problem 4361 and Bro-card’s theorem that the center of a circle is the orthocenter of triangles that areself-polar with respect to the circle (meaning each side of the triangle is the poleof the opposite vertex): ∆PQH ′ is an appropriate self-polar triangle with respectto Γ, therefore its orthocenter is O (which implies, of course, that H ′ is the ortho-center of ∆OPQ). Bataille observed that the polar of H ′, namely the line PQ,contains the incenter I. That can easily be confirmed using the coordinates of ourfeatured solution: subtract the coordinates of Q from those of P and you get thecoordinates of I(a : b : c).

4362. Proposed by Oai Thanh Dao and Leonard Giugiuc.

Let ABCD be a convex quadrilateral and let F be the midpoint of CD. Considera point E inside ABCD such that AE · CE = BE ·DE. The lines EF and ABintersect at G. If ∠AED + ∠CEB = 180◦, prove that ∠AED = ∠AGE.

We received 4 solutions and will feature just one of them here, by Michel Bataille.

Let lines m and n pass through E and be parallel and perpendicular to AB,respectively. Let M on m and N on n be such that EM = EN = 1. Consider mas the x-axis and n as the y-axis of a system of axes with origin at E. Then toeach point is assigned a complex affix. If y0 denotes the distance from E to theline AB, the affixes a and b of A and B are a = x1 + iy0 and b = x2 + iy0 for somereal numbers x1, x2. Let α = ∠AED. Then α ∈ (0, π) and we may suppose that

M and N are chosen such that the oriented angle ∠(−→EA,−−→ED) equals α, and then,

by assumption ∠(−−→EC,

−−→EB) = π − α (see the figure on the next page).


Solutions /285

Let c and d denote the affixes of C and D. Since AE · CE = BE ·DE, we have∣∣∣∣da∣∣∣∣ =

∣∣∣cb

∣∣∣ = ρ, say, so that

d

a= ρeiα and

b

c=

1

ρei(π−α).

We deducec

b= −ρeiα

and the affix f of F is given by

f =c+ d

2= ρeiα(a− b) = ρeiα(x1 − x2).

From the equation

e−iαz − eiαz = 0

of EF and z − z = 2iy0 of AB, we readily obtain the affix g of G:

g =y0e

iα

sinα.

Observing that y0 > 0 and sinα > 0, we see that α is an argument of g, which

means that ∠(−−→EM,

−−→EG) = α. Since AG ‖ EM , it follows that ∠(

−→GA,−−→GE) = α

and so ∠AGE = ∠ADE, as required.


286/ Solutions


Let (an)n≥0 be the sequence defined by a0 > 0 and the recursion

an+1 =an

1 + (n+ 1)a2n.

Prove that the series∞∑n=0

a2n is convergent and find limn→∞

(n ·

∞∑k=n

a2k

).

We received 7 solutions. We present the solution by Oliver Geupel, lightly edited.

The sequence bn = (n+ 1)an, where n ≥ 0, satisfies the recursion

bn+1 = (n+ 2)bn/(n+ 1 + b2n).

Note that bn > 0 for all n. We show by induction that bn ≤ 1 for n ≥ 1. First, wehave b1 = 2b0/(1 + b20) ≤ 1, which is the base case. Assuming the assertion holdsfor some index n, we obtain

bn+1 =nbn + 2bnn+ 1 + b2n

≤ n+ (1 + b2n)

n+ 1 + b2n= 1,

and the induction is complete. Hence an ≤ 1n+1 for n ≥ 1, and we obtain

∞∑n=0

a2n ≤ a20 +

∞∑n=1

1

(n+ 1)2= a20 +

π2

6− 1;

that is,∞∑n=0

a2n converges.

For n ≥ 1, note that

bn+1 =(n+ 2)bnn+ 1 + b2n

≥ (n+ 2)bnn+ 2

= bn.

The sequence (bn)n≥1 is thus increasing and bounded above by 1, which impliesthat it has a limit 0 < L ≤ 1. From the recursion formula for bn+1 (taking the

limit as n → ∞ of both sides) we get L =(n+ 2)L

n+ 1 + L2. The only positive root of

this equation is L = 1, which must be the desired limit.

Note that

n ·∞∑k=n

a2k = n ·∞∑k=n

b2k(k + 1)2

.

For every ε > 0 there is an index n0 such that, whenever n > n0, we have1− ε < bn ≤ 1. Moreover, for n > 0,

1

n+ 1=

∫ ∞n+1

x−2 dx <

∞∑k=n

1

(k + 1)2<

∫ ∞n

x−2 dx =1

n.


Solutions /287

Hence for n > n0,

n

n+ 1· (1− ε)2 < n ·

∞∑k=n

a2k ≤ 1,

and so limn→∞

(n ·

∞∑k=n

a2k

)= 1.


Let f : N×N→ N be a binary operation, and define g : N×N→ N by

g(a, b) =

{a if b = 1f(g(a, b− 1), a) if b ≥ 2.

If f is associative and g is commutative, prove that f(a, b) = a+b and g(a, b) = ab.

We received 6 solutions. We present the solution by the Missouri State UniversityProblem Solving Group.

Note that, since g is commutative, from the definition of g it follows that g(1, a) = afor all a ∈ N.

Let n ∈ N. Then

n+ 1 = g(1, n+ 1) = f(g(1, n), 1) = f(n, 1).

That is,

f(n, 1) = n+ 1 (1)

for every n ∈ N.We now show that f(n,m) = n+m. Fix n and use induction on m ≥ 1.

f(n,m+ 1) = f(n, f(m, 1)) (by (1), the base case)

= f(f(n,m), 1) (by associativity of f)

= f(n,m) + 1 (by (1))

= n+m+ 1 (by the induction hypothesis).

Next we show g(n,m) = nm. Once again fix n and use induction on m ≥ 1. Notethat g(n, 1) = n · 1 by the definition of g.

g(n,m+ 1) = f(g(n,m), n) (by the definition of g)

= f(nm, n) (by the induction hypothesis)

= nm+ n (by the formula for f proved above)

= n(m+ 1).

This completes the proof.


288/ Solutions

4365. Proposed by Marius Dragan and Neculai Stanciu.

Let a and b be real numbers such that a + b, a4 and b4 are rational numbers anda+ b 6= 0. Prove that a and b are rational numbers.

We received 7 solutions and will feature the solution by Madhav Modak.

Let a+ b = r, a4 = c and b4 = d. Then by assumption, r, c, d are rational numbersand r 6= 0. Eliminating b from a+ b = r and b4 = d, we get (r − a)4 = d or

a4 − 4ra3 + 6r2a2 − 4r3a+ r4 = d.

Hence− 4ar(a2 + r2) = t− 6a2r2, where t = d− r4 − c. (1)

Squaring (1), we get:

16a2r2(c+ r4 + 2a2r2) = t2 + 36cr4 − 12a2r2t,

a2(16cr2 + 16r6) + 32cr4 = t2 + 36cr4 − 12a2r2t,

a2(16cr2 + 16r6 + 12r2t) = t2 + 4cr4,

a2(16cr2 + 16r6 + 12r2d− 12r6 − 12cr2) = t2 + 4cr4,

a2(4cr2 + 4r6 + 12r2d) = t2 + 4cr4. (2)

Now r 6= 0 so that r6 > 0 and so 4cr2 + 4r6 + 12r2d > 0 as c, d ≥ 0. Hence by (2),

a2 =t2 + 4cr4

4cr2 + 4r6 + 12r2d,

so that a2 is rational. Then by (1),

a = (6a2r2 − t)/4r(a2 + r2),

as 4r(a2 + r2) 6= 0, we see that a is rational, hence b = r− a is rational, as was tobe shown.

4366. Proposed by Daniel Sitaru.

Let xn be the base angle of a right triangle with base n and altitude 1. Find

∞∑k=1

xk2+k+1.

There were 15 correct solutions, all variants of the following two solutions.

Solution 1.

The arms of the right triangle have lengths 1 and n, and xn is the angle adjacentto the latter arm. Thus, xn = arctan 1

n . Observe that

tan(xk − xk+1) =1k − 1

k+1

1 + 1k(k+1)

=1

k2 + k + 1= tan(xk2+k+1).


Solutions /289

Therefore

n∑k=1

xk2+k+1 =

n∑k=1

(xk − xk+1)

= x1 − xn+1

=π

4− arctan

1

n+ 1,

so that∞∑k=1

xk2+k+1 =π

4.

Solution 2.

Let

un = tan

(n∑k=1

xk2+k+1

).

Checking the values of un for small values of n, we are led to the conjecture thatun = n(n+ 2). Suppose that this holds for n = m− 1. Then

un = tan

(xm2+m+1 +

m−1∑k=1

xk2+k+1

)

=

1

m2 +m+ 1+m− 1

m+ 1

1− m− 1

(m+ 1)(m2 +m+ 1)

=m(m2 + 1)

(m+ 2)(m2 + 1)=

m

m+ 2.

Thus, an induction argument, along with u1 = 1/3, establishes that un = n/(n+2)for each positive integer n. Since the limit as n tends to infinity of un is 1, thesum of the given series is π

4 .

4367. Proposed by Kadir Altintas and Leonard Giugiuc.

Let a, b and c be distinct complex numbers such that |a| = |b| = |c| = 1 and|a+ b+ c| ≤ 1. Prove that∣∣∣∣ (a+ b)(b+ c)

(a− b)(b− c)

∣∣∣∣+

∣∣∣∣ (b+ c)(c+ a)

(b− c)(c− a)

∣∣∣∣+

∣∣∣∣ (c+ a)(a+ b)

(c− a)(a− b)

∣∣∣∣ = 1.

We received 3 correct solutions. We present the solution by Michel Bataille.

Let u =(c+ a)(a+ b)

(c− a)(a− b) , v =(a+ b)(b+ c)

(a− b)(b− c) , w =(b+ c)(c+ a)

(b− c)(c− a).

If for example b+ c = 0, then v = w = 0 and u = −1, hence |u|+ |v|+ |w| = 1 issatisfied.


290/ Solutions

In what follows, we suppose that b+ c, c+ a, a+ b are nonzero complex numbers.

Since the conjugate of a, b, c are 1a ,

1b ,

1c , respectively, an easy calculation gives

u = u, v = v, w = w, meaning that u, v, w are real numbers.

Similarly, z =(a+ b)(b+ c)(c+ a)

(a− b)(b− c)(c− a)is a nonzero purely imaginary number and so

uvw = z2 is a negative real number.

The three following identities are readily checked:

(a+b)(b+c)(c−a)+(b+c)(c+a)(a−b)+(c+a)(a+b)(b−c) = −(a−b)(b−c)(c−a)(1)

(a+b)(b−c)(c−a)+(b+c)(c−a)(a−b)+(c+a)(a−b)(b−c) = 8abc−(a+b)(b+c)(c+a)(2)

(a+ b+ c)(ab+ bc+ ca) = (a+ b)(b+ c)(c+ a) + abc. (3)

Identity (1) shows that u+ v + w = −1 while (2) easily gives

uv + vw + wu = z2(

8abc

(a+ b)(b+ c)(c+ a)− 1

). (4)

But the condition |a+ b+ c| ≤ 1 writes as

(a+ b+ c)

(1

a+

1

b+

1

c

)≤ 1,

that is,(a+ b)(b+ c)(c+ a) + abc

abc≤ 1

(using (3)) or finally(a+ b)(b+ c)(c+ a)

abc≤ 0.

From this and (4), we deduce that uv + vw + wu > 0.

Now, let

p(x) = (x− u)(x− v)(x− w) = x3 + x2 + (uv + vw + wu)x− uvw.

The derivativep′(x) = 3x2 + 2x+ (uv + vw + wu)

is positive on [0,∞) (since uv + vw + wu > 0), hence p is an increasing functionon [0,∞) and therefore

p(x) > p(0) = −uvw > 0

for x ∈ [0,∞). Thus, the roots u, v, w of p(x) are negative real numbers and so

|u|+ |v|+ |w| = −u− v − w = −(u+ v + w) = −(−1) = 1,

as desired.


Solutions /291

4368. Proposed by Ovidiu Furdui and Alina Sıntamarian.

Calculate∞∑n=2

[2n (ζ(n)− 1)− 1] ,

where ζ denotes the Riemann zeta function defined as ζ(z) =∞∑n=1

1

nz.

We received 9 submissions, all correct, although one used WolframAlpha software.We present the solution by Paolo Perfetti, which is short, simple, and similar tothe other seven non-WolframAlpha solutions.

∞∑n=2

[2n(ζ(n)− 1)− 1] =

∞∑n=2

[2n∞∑k=2

1

kn− 1

]

=

∞∑n=2

∞∑k=3

2n

kn

=

∞∑k=3

∞∑n=2

2n

kn

=

∞∑k=3

4

k21

1− 2k

= 2

∞∑k=3

[1

k − 2− 1

k − 1+

1

k − 1− 1

k

]=︸︷︷︸

telescoping

2(1 +1

2) = 3.

4369. Proposed by Mihaela Berindeanu.

On the sides of triangle ABC, take points

A1, A2 ∈ (BC) , B1, B2 ∈ (AC) , C1, C2 ∈ (AB) ,

so thatBA1 = A2C, CB1 = B2A, AC1 = C2B.

On B2C1, A1C2, A2B1 take A3, B3, C3 so that

C1A3

A3B2=A1B3

B3C2=B1C3

C3A2= k.

Find all values of k for which AA3, BB3, CC3 are concurrent lines.

We received 5 correct submissions. We present the solution by Michel Bataille.


292/ Solutions

Let α > 0 be such that−−→BA1 = α

−−→BC =

−−→A2C. Then,

A1 = (1− α)B + αC and A2 = αB + (1− α)C.

In a similar way, there exist β, γ > 0 such that

B1 = βA+ (1− β)C,

B2 = (1− β)A+ βC,

C1 = (1− γ)A+ γB,

C2 = γA+ (1− γ)B.

Expressing that−−−→C1A3 = k

−−−→A3B2, we obtain (1 + k)A3 = kB2 + C1, that is,

(1 + k)A3 = ρA+ γB + kβC,

where ρ = 1 + k − γ − kβ. Similarly,

(1 + k)B3 = kγA+ σB + αC,

(1 + k)C3 = βA+ kαB + τC,

where σ, τ ∈ R. It readily follows that the equations of the lines AA3, BB3, CC3

respectively are

kβy − γz = 0, αx− kγz = 0, kαx− βy = 0.

Now, the common point of BB3 and CC3 is J = (kβγ : k2αγ : αβ) (not at infinitysince kβγ + k2αγ + αβ is positive, hence not zero), and this point J is on AA3 ifand only if kβ(k2αγ)− γ(αβ) = 0, that is, if and only if k3 = 1.

We conclude that AA3, BB3, CC3 are concurrent if and only if k = 1 (whichmeans that A3, B3, C3 are the midpoints of B2C1, C2A1, A2B1, respectively).

4370. Proposed by Leonard Giugiuc and Sladjan Stankovik.

Solve the following system of equations:a+ b+ c+ d = 4,

a2 + b2 + c2 + d2 = 7,

abc+ abd+ acd+ bcd− abcd = 1516 .

We received 10 submissions, including the one from the proposers, 9 of which arecorrect, and the other one, incomplete. We present the same solution by DionneBailey, Elsie Campbell, Charles Diminnie, and Trey Smith (jointly); and Ramanu-jan Srihari.

Note first that

16 = (a+ b+ c+ d)2

= a2 + b2 + c2 + d2 + 2(ab+ ac+ ad+ bc+ bd+ cd)

= 7 + 2(ab+ ac+ ad+ bc+ bd+ cd)


Solutions /293

so

ab+ ac+ ad+ bc+ bd+ cd =9

2.

Letting w = a− 1, x = b− 1, y = c− 1, and z = d− 1, we then have

w + x+ y + z = 0

and

wxyz = (a− 1)(b− 1)(c− 1)(d− 1)

= (abcd)− (abc+ abd+ acd+ bcd)

+ (ab+ ac+ ad+ bc+ bd+ cd)− (a+ b+ c+ d) + 1

= −15

16+

9

2− 4 + 1

=9

16.

Also,

w2 + x2 + y2 + z2 = (a− 1)2 + (b− 1)2 + (c− 1)2 + (d− 1)2

= a2 + b2 + c2 + d2 − 2(a+ b+ c+ d) + 4

= 7− 8 + 4

= 3.

By the AM-GM Inequality, we have

3 = w2 + x2 + y2 + z2 ≥ 4 4√w2x2y2z2 = 4

√9

16= 3

with equality if and only if

w2 = x2 = y2 = z2 =3

4.

Hence,

(w, x, y, z) =(±√

3

2,±√

3

2,±√

3

2,±√

3

2

).

Since w+ x+ y+ z = 0, we then easily see that exactly two of w, x, y, and z must

be√32 , and the other two equal -

√32 . Therefore, there are six solutions given by

(a, b, c, d) = (1 +

√3

2, 1 +

√3

2, 1−

√3

2, 1−

√3

2)

together with its five permutations.


294/ Solutions

4371. Proposed by Oai Thanh Dao and Leonard Giugiuc.

Let two perpendicular lines pass through the orthocenter H of a triangle ABC.Suppose they meet the sides AB and AC in C1, B1 and C2, B2, respectively. DefineMi as the midpoint of BiCi (i = 1, 2) and M as the midpoint of BC. Prove thatM,M1 and M2 are collinear.

We received 6 submissions, of which 5 were complete and correct. We present thesolution by Michel Bataille.

The perpendicular lines through H define a system of orthonormal axes with originat H. We assign the coordinates

B1(2b1, 0), C1(2c1, 0), B2(0, 2b2), C2(0, 2c2)

for some real numbers b1, b2, c1, c2. Then, M1(b1 + c1, 0), M2(0, b2 + c2) and theequation of the line M1M2 is

(b2 + c2)x+ (b1 + c1)y = (b1 + c1)(b2 + c2). (1)

The equation of the line AB = C1C2 is c2x+ c1y = 2c1c2. The altitude from B isperpendicular to AC = B1B2 and passes through the origin H, so its equation isb1x− b2y = 0. The lines AB and BH intersect in B, so we calculate

B

(2c1c2b2

b1c1 + b2c2,

2c1c2b1b1c1 + b2c2

).

Similarly (mutatis mutandis), we have

C

(2b1b2c2

b1c1 + b2c2,

2b1b2c1b1c1 + b2c2

).


Solutions /295

Thus the midpoint M of BC has coordinates

xM =b2c2(b1 + c1)

b1c1 + b2c2and yM =

b1c1(b2 + c2)

b1c1 + b2c2;

and

(b2 + c2)xM + (b1 + c1)yM =b2c2(b1 + c1)(b2 + c2)

b1c1 + b2c2+b1c1(b1 + c1)(b2 + c2)

b1c1 + b2c2= (b1 + c1)(b2 + c2)

and from (1) we obtain that M is on the line M1M2, as desired.

4372. Proposed by J. Chris Fisher.

Given a quadrangle ABCD with right angles at B and D, let the circle (D,DA)(with center D and radius DA) intersect the line AC at P and the circle (B,BA)at Q. Prove that PQ is perpendicular to AB.

We received 7 submissions, of which 5 were correct and complete. We presenta solution based on the submission by Sushanth Sathish Kumar, completed andcorrected by the editor.

Since AQ is the radical axis of circles (B,BA) and (D,DA), the line DB bisects∠QDA. By the inscribed angle theorem,

∠QPA =360◦ − ∠QDA

2= 180◦ − ∠BDA.

Since ∠QPC and ∠QPA are supplementary, it follows that ∠QPC = ∠BDA.Quadrilateral ABCD is cyclic, so also ∠BDA = ∠ACB. Therefore, we have∠QPC = ∠PCB (since ∠PCB and ∠ACB are the same angle), which impliesPQ ‖ CB. From this we can conclude PQ ⊥ AB, as desired.


296/ Solutions


Let p be an odd prime. Let q and r be the quotient and the remainder in the

division of the positive integer n by p and let Sn =n∑k=r

(−1)k−r(kr

). Show that

Sn ≡ 0 (mod p) if and only if q is odd and 2r ≡ 1 (mod p).

We received three submissions, including the one from the proposer. We presentthe solution by Modhav R. Modak, modified slightly by the editor.

By assumption, we have n = qp+ r. For i = 0, 1, 2, . . . , q − 1, let

Ai =

p−1∑k=0

(−1)k(r + ip+ k

r

).

Since p is odd, (−1)ip = 1 or −1 depending on whether i is even or odd. Hencewe can write

Sn =

p−1∑k=0

(−1)k(r + k

r

)+

2p−1∑k=p

(−1)k(r + k

r

)+ · · ·+

qp−1∑k=q

(−1)k(r + k

r

)

= A0 −A1 + · · ·+ (−1)qp−1Aq−1 + (−1)qp(r + qp

r

). (1)

Since p > 2 and r < p, we have for r ≤ k,

(k + p)(k + p− 1) · · · (k + p− r + 1) ≡ (k)(k − 1) · · · (k − r + 1) (mod p)

so that

(k + p

r

)≡(k

r

)(mod p) since p 6 |r!. (2)

By (2) we then have, for i = 0, 1, 2, . . . , q − 1, that

Ai ≡(r

r

)−(r + 1

r

)+ · · ·+

(r + p− 1

r

)(mod p)

i.e. Ai ≡ A0 (mod p) .

Hence from (1) we get, modulo p,

Sn ≡{

1, if q is even,

A0 − 1 if q is odd.(3)

We next establish the following identity:

m∑k=0

(−1)k(r + k

r

)=

1

2r+1

[1 + (−1)m

{r∑

k=0

2k(m+ k

k

)}], (4)

To prove (4), note that

∞∑k=0

(−1)k(m+ k

k

)xk =

1

(1 + x)m+1.


Solutions /297

Hence

m∑k=0

(−1)k(r + k

r

)= coefficient of xm in f(x) =

1

(1− x)(1 + x)m+1.

Now f(x) can be decomposed into partial fractions as

f(x) =1

2r+1

[1

1− x +1

1 + x+

2

(1 + x)2+

22

(1 + x)3+ · · ·+ 2r

(1 + x)r+1

](5)

since the right hand side of (5) can be written as

=1

2r+1

[1

1− x +1

2

r+1∑k=1

2k

(1 + x)k

]

=1

2r+1

[1

1− x +1

2· 2

1 + x· (2/(1 + x))r+1 − 1

(2/(1 + x))− 1

]

=1

2r+1

[1

1− x +2r+1 − (1 + x)r+1

(1− x)(1 + x)r+1

]= f(x).

Hence (4) follows by taking the coefficient of xm in (5).

Letting m = p− 1 in (4), we then have

A0 =

p−1∑k=0

(−1)k(r + k

r

)=

1

2r+1

(1 + (−1)p−1

(p−1∑k=0

2k(p− 1 + k

k

)))

=1

2r+1

(1 + 1 + 2

(p

1

)+ 22

(p+ 1

2

)+ · · ·+ 2p−1

(2p− 2

p− 1

)),

(6)

so

A0 ≡1

2r(mod p) ,

since (p

1

),

(p+ 1

2

), · · · ,

(2p− 2

p− 1

)are all congruent to zero modulo p.

Hence, (3) becomes

Sn ≡

1, if q is even,1

2r· (1− 2r), if q is odd.

(7)

Thus if q is odd and 2r ≡ 1 (mod p) , then (6) shows that Sn ≡ 0 (mod p) .Further, if q is even or if 2r 6≡ 1 (mod p) , then Sn 6≡ 0 (mod p) , completing theproof.


298/ Solutions

4374. Proposed by Sefket Arslanagic.

For a fixed positive integer n, solve the equation

|1− |2− |3− · · · − | n− x | · · · · · · |︸︷︷︸n vertical bars

= 1.

There were 6 correct solutions. We present the standard approach.

Let En denote the given equation, Sn the set of its solutions, and Tn = 12n(n+ 1),

the sum of the first n positive integers. It is readily checked that S1 = {0, 2},S2 = {0, 2, 4}, S3 = {−1, 1, 3, 5, 7}. We prove by induction that

Sn = {−Tn−2,−Tn−2 + 2,−Tn−2 + 4, . . . ,−Tn−2 + 2(Tn−1 + 1) = Tn + 1}

for n ≥ 3.

Suppose that this holds for n = m. Then x is a solution of Em+1 if and only if|m+ 1− x| is a nonnegative solution of Em, if and only if

|m+ 1− x| = −Tm−2 + 2k

for 12Tm−2 ≤ k ≤ Tm−1 + 1.

The smallest value assumed by |m + 1 − x| for a solution x is equal to 0 whenm ≡ 1, 2 (mod 4) and 1 when m ≡ 0, 3 (mod 4). It follows that the numbers inSm+1 are all consecutively even when m ≡ 0, 1 (mod 4) and all consecutively oddwhen m ≡ 2, 3 (mod 4).

The solution x assumes its smallest value when

x = (m+ 1)− (Tm + 1) = −(Tm −m) = −Tm−1

and its largest value when

x = (m+ 1) + (Tm + 1) = Tm+1 + 1.

Thus, Sm+1 consists of all integers between −Tm−1 and Tm+1 + 1 inclusive thatshare the same parity.


Consider two sequences {xn} and {yn} such that {xn} ∪ {yn} = N. Let l > 1 begiven. Prove that lim inf xn/n ≥ l if and only if lim sup yn/n ≤ l/(l − 1).

The problem is incorrect as it stands. A counterexample was offered by OliverGeupel: let xn = n2, y2n−1 = n2 and y2n = n. If you want one where thetwo sequences are exhaustive but not overlapping take xn = 3n and y2m = 2m+1

with the other yn picking up the missing numbers. In this case, take l = 3 andl/(l − 1) = 3/2.


Solutions /299

The following solution is essentially that of the proposer, with additional hypothesesto make it work.

We assume that both {xn} and {yn} are increasing, nonoverlapping and jointlyexhaustive sequences of positive integers. Let t be a positive integer and let N(t)be the number of indices n for which yn ≤ t and M(t) the number of indices n forwhich xn ≤ t. Then M(t) +N(t) = t.

Suppose xm ≤ t < xm+1. Then M(t) = m and

xmm≤ t

M(t)<

(xm+1

m+ 1

)(m+ 1

m

)so that

lim inft→∞

t

M(t)= lim inf

m→∞

xmm

= a ≥ 1.

Similarly

lim supt→∞

t

N(t)= lim sup

m→∞

ymm

= b ≥ 1.

SinceM(t)

t+N(t)

t= 1

and lim supM(t)/t = 1− lim inf N(t)/t, we find that 1a + 1

b = 1 and b = aa−1 .

Let l be as in the hypothesis. Then,

lim inf xn/n = lim inf t/M(t) ≥ l⇔ lim supM(t)/t ≤ 1/l

⇔ lim inf N(t)/t ≥ 1− (1/l) = (l − 1)/l

⇔ lim sup t/N(t) = lim sup yn/n ≤ l/(l − 1).


Documents

SOLUTIONS - cms.math.ca · 270/ Solutions SOLUTIONS No problem is ever permanently closed. The editor is always pleased to consider for publication new solutions or new insights on