solutions...show that for any triangle abc, the following inequality holds sinasinbsinc † 1 sina+...

SOLUTIONS /31

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 2015: 41(1), p. 27–30.

4001. Proposed by Cristinel Mortici and Leonard Giugiuc.

Let a, b, c, d ∈ R with d > 2 such that

(2d+ 1) · a6

+b

2+

c

d+ 1= 0.

Prove that there exists t ∈ (0, d) such that at2 + bt+ c = 0.

We received four correct solutions and one solution that was almost complete. Thefirst solution is due to Digby Smith and the second consists of ingredients of others.

Solution 1, by Digby Smith.

The result actually holds when d > 1. Let

2v = d+ 1 and 3w = 2d+ 1.

Then 0 < v < w < d with 6(w− v) = d− 1. The given condition can be rewrittenas

0 = avw + bv + c = v(aw + b) + c.

Let f(t) = at2 + bt+ c. Then

f(w) = aw2 + bw + c = w(aw + b) + c = −wcv

+ c = −c(w − v)

v= −c(d− 1)

6v.

If c 6= 0, then it follows that f(0) = c and f(w) have opposite signs, so that f(t)has a real root in the interval (0, w) ⊆ (0, d).

If c = 0, then f(w) = w(aw + b) = 0 since v(aw + b) = 0.

Solution 2.

Use the notation of Solution 1. Again 0 < v < w < d. Furthermore, whena = 0, f(v) = 0. Otherwise, we may assume that a > 0, in which case f(v) <avw + bv + c = 0.

When c = 0, then f(w) = 0. When c > 0, then f(0) > 0 and f(t) has a root in(0, v). Finally, when c < 0, then avw + bv = −c > 0 and

f(w) = aw2 + bw − avw − bv = (w − v)(aw + b) > 0

and f(t) has a root in (v, w).

Copyright c© Canadian Mathematical Society, 2016

32/ SOLUTIONS

4002. Proposed by Henry Aniobi.

Let f be a convex function on an interval I. Let x1 ≤ x2 ≤ . . . ≤ xn andy1 ≤ y2 ≤ . . . ≤ yn be numbers such that xi + yj is always in I for all 1 ≤ i, j ≤ n.Let z1, z2, . . . , zn be an arbitrary permutation of y1, y2, . . . , yn. Show that

f(x1 + y1) + . . .+ f(xn + yn) ≥ f(x1 + z1) + . . .+ f(xn + zn)

≥ f(x1 + yn) + f(x2 + yn−1) + . . .+ f(xn + y1);

We received five submissions of which four were correct and complete. We presentthe solution by Joseph DiMuro.

We can prove the above statement by proving the following simpler statement:

Claim. Let x1 < x2 and y1 < y2 be numbers such that xi+yj is always in I. Then

f(x1 + y1) + f(x2 + y2) ≥ f(x1 + y2) + f(x2 + y1).

The reason why this suffices: if we choose a permutation z1, z2, . . . , zn such thatzi > zj for some i < j, then we will have

f(xi + zi) + f(xj + zj) ≤ f(xi + zj) + f(xj + zi).

We would then be able to interchange zi and zj without decreasing the overallsum. Thus, a permutation z1, z2, . . . , zn that gives us the largest overall sum isone where zi ≤ zj whenever i < j; that is, zi = yi for all i. Similarly, a permutationz1, z2, . . . , zn that gives us the smallest overall sum is one where zi ≥ zj wheneveri < j; that is, zi = yn−i+1 for all i.

Proof of claim. By the definition of convexity, for all a, b ∈ I and for all t ∈ [0, 1],we have

f(ta+ (1− t)b) ≤ tf(a) + (1− t)f(b).

Let a = x1 + y1 and b = x2 + y2. Then a < x1 + y2 < b, so for some t ∈ [0, 1], wehave x1 + y2 = ta+ (1− t)b. From that, we have:

x2 + y1 = (x1 + y1 + x2 + y2)− (x1 + y2)

= (a+ b)− (ta+ (1− t)b)= (1− t)a+ tb.

Therefore,

f(x1 + y2) + f(x2 + y1) = f(ta+ (1− t)b) + f((1− t)a+ tb)

≤ (tf(a) + (1− t)f(b)) + ((1− t)f(a) + tf(b))

= f(a) + f(b)

= f(x1 + y1) + f(x2 + y2),

completing the proof of the claim.

Crux Mathematicorum, Vol. 42(1), January 2016

SOLUTIONS /33

4003. Proposed by Martin Lukarevski.

Show that for any triangle ABC, the following inequality holds

sinA sinB sinC

Å1

sinA+ sinB+

1

sinB + sinC+

1

sinC + sinA

ã≤ 3

4(cosA+ cosB + cosC).

We received 13 correct solutions. We present the solution by John G. Heuver,modified slightly by the editor.

Let r,R and s denote the inradius, the circumradius and the semiperimeter of4ABC, respectively. The following identities and inequalities are well known:∑

sin2A =s2 − 4Rr − r2

2R2, (1)∑

sinA sinB =s2 + 4Rr + r2

4R2, (2)∑

cosA =R+ r

R, (3)

R ≥ 2r Euler’s inequality (4)

s2 ≤ 4R2 + 4Rr + 3r2 Gerretsen’s inequality, (5)

(where all the summations are taken over all angles of 4ABC).

Let L denote the left-hand side of the given inequality. By the AM-GM Inequality,we have sinA + sinB ≥ 2

√sinA sinB. Hence, by Cauchy-Schwarz Inequality we

have

L ≤ 1

2

∑(sinA)

√sinB sinC ≤ 1

2

√∑sin2A

√∑sinA sinB. (6)

By (1) and (5), we have ∑sin2A ≤ 4R2 + 2r2

2R2. (7)

By (2) and (5), we have

∑sinA sinB ≤ 4R2 + 8Rr + 4r2

4R2. (8)

Using (6), (7) and (8) followed by (3) and (4), we then have

L ≤ 1

2

…4R2 + 2r2

2R2·…

4R2 + 8Rr + 4r2

4R2=

1

2

…2 +

( rR

)2· R+ r

R

≤ 1

2

2 +

Å1

2

ã2·∑

cosA

=3

4

∑cosA,


34/ SOLUTIONS

which completes the proof.

Editor’s comment. Digby Smith remembered that the following problem proposedby Jack Garfunkel and George Tsintsifas appeared in the August–September 1982issue (Vol. 8, no. 7, p. 210) of Crux and a solution given by Vedula N. Murtyappeared in the November 1983 issue (Vol. 9, no. 9, p. 282):

4

9

∑sinB sinC ≤

∏cos

B − C2

≤ 2

3

∑cosA.

Smith gave a proof by first showing that 2L ≤∑

sinB sinC, which together withthe above inequality yields the result.

4004. Proposed by George Apostolopoulos.

Let x, y, z be positive real numbers such that x+ y + z = 2. Prove that

x5

yz(x2 + y2)+

y5

zx(y2 + z2)+

z5

xy(z2 + x2)≥ 1.

We received 16 correct submissions. We present 3 solutions.

Solution 1, by Arkady Alt.

Since by Cauchy’s Inequality

∑cyc

x5

yz (x2 + y2)=∑cyc

x6

xyz (x2 + y2)≥(x3 + y3 + z3

)2∑cyc

xyz (x2 + y2),

it suffices to prove the inequality(x3 + y3 + z3

)2∑cyc

xyz (x2 + y2)≥ 1.

We have the following equivalences:(x3 + y3 + z3

)2∑cyc

xyz (x2 + y2)≥ 1 ⇐⇒

(x3 + y3 + z3

)2 ≥ 2xyz(x2 + y2 + z2

)⇐⇒

(x3 + y3 + z3

)2 ≥ xyz (x+ y + z)(x2 + y2 + z2

),

where the latter inequality holds because by AM-GM Inequality

x3 + y3 + z3 ≥ 3xyz

and by Chebyshev’s Inequality

x3 + y3 + z3 ≥(x+ y + z)

(x2 + y2 + z2

)3

.


SOLUTIONS /35

Solution 2, by Michel Bataille.

Let a = x2 , b = y

2 and c = z2 . With these notations, we are required to prove

a6

a2 + b2+

b6

b2 + c2+

c6

c2 + a2≥ abc

2(1)

under the conditions a, b, c > 0 and a+ b+ c = 1.

The Cauchy-Schwarz inequality givesÅa6

a2 + b2+

b6

b2 + c2+

c6

c2 + a2

ã((a2 + b2) + (b2 + c2) + (c2 +a2)) ≥ (a3 + b3 + c3)2.

Hence, (1) will follow if we prove

(a3 + b3 + c3)2

a2 + b2 + c2≥ abc.

Since abc ≤ a3+b3+c3

3 , it is sufficient to show that

3(a3 + b3 + c3) ≥ a2 + b2 + c2.

Now, the latter follows from

3(a3 + b3 + c3) ≥ 2(a3 + b3 + c3) + 3abc

= a3 + b3 + c3 + (a3 + b3 + c3 + 3abc)

≥ a3 + b3 + c3 + ab2 + a2b+ bc2 + b2c+ ca2 + c2a (Schur’s ineq.)

= (a2 + b2 + c2)(a+ b+ c) = a2 + b2 + c2 (since a+ b+ c = 1)

so we are done.

Solution 3, by Oliver Geupel.

By hypothesis x+ y + z = 2 and by the Cauchy-Schwarz inequality we haveÅx5

yz(x2 + y2)+

y5

zx(y2 + z2)+

z5

xy(z2 + x2)

ã(x+ y + z)xyz(x2 + y2 + z2)

=

(∑cyc

x5

yz(x2 + y2)

)(∑cyc

xyz(x2 + y2)

)≥ (x3 + y3 + z3)2.

By the power mean inequality, it holdsÅx3 + y3 + z3

3

ã1/3≥Åx2 + y2 + z2

3

ã1/2≥ x+ y + z

3≥ (xyz)1/3.


36/ SOLUTIONS

Putting together we obtain

x5

yz(x2 + y2)+

y5

zx(y2 + z2)+

z5

xy(z2 + x2)

≥ (x3 + y3 + z3)2

(x+ y + z)xyz(x2 + y2 + z2)

=(x3 + y3 + z3)1/3

x+ y + z· x

3 + y3 + z3

xyz· (x3 + y3 + z3)2/3

x2 + y2 + z2

≥ 3−2/3 · 3 · 3−1/3 = 1.

Hence the result. By the equality condition of the power mean inequality, theequality holds if and only if x = y = z = 2/3.

4005. Proposed by Michel Bataille.

Let a, b, c be the sides of a triangle with area F . Suppose that some positive realnumbers x, y, z satisfy the equations

x+ y + z = 4 and

2xb2c2 + 2yc2a2 + 2za2b2 −Å

4− yzx

a4 +4− zxy

b4 +4− xyz

c4ã

= 16F 2.

Show that the triangle is acute and find x, y, z.

We present the proposer’s solution — no others were submitted.

The second equation gives

(xyz)(16F 2)

= xyz(2xb2c2 + 2yc2a2 + 2za2b2)− yz(4− yz)a4 − zx(4− zx)b4 − xy(4− xy)c4

= (a2yz + b2zx+ c2xy)2 − (4a4yz + 4b4zx+ 4c4xy)

=(x

2(b2z + c2y) +

y

2(c2x+ a2z) +

z

2(b2x+ a2y)

)2− (4a4yz + 4b4zx+ 4c4xy).

Since t 7→ t2 is a convex function and x+ y + z = 4, Jensen’s inequality yields

x

4(b2z + c2y)2 +

y

4(c2x+ a2z)2 +

z

4(b2x+ a2y)2

≥(x

4(b2z + c2y) +

y

4(c2x+ a2z) +

z

4(b2x+ a2y)

)2(1)

and it follows that

(xyz)(16F 2)

≤ x(b2z + c2y)2 + y(c2x+ a2z)2 + z(b2x+ a2y)2 − (4a4yz + 4b4zx+ 4c4xy)

= a4yz(y + z − 4) + b4zx(z + x− 4) + c4xy(x+ y − 4) + xyz(2b2c2 + 2c2a2 + 2a2b2)

= xyz(2b2c2 + 2c2a2 + 2a2b2 − a4 − b4 − c4)

= (xyz)(16F 2).


SOLUTIONS /37

Thus, equality must hold in (1) and because t 7→ t2 is a strictly convex function,this calls for

b2z + c2y = c2x+ a2z = b2x+ a2y.

Setting these three expressions equal to λ and solving for x, y, z yields

x = λb2 + c2 − a2

2b2c2=λ cosA

bc,

y = λc2 + a2 − b2

2c2a2=λ cosB

ca,

z = λa2 + b2 − c2

2a2b2=λ cosC

ab.

(As usual, A,B,C denote the angles of the triangle opposite sides a, b, c, respec-tively.) Since at most one of A,B,C is not acute and x, y, z are positive, weconclude that cosA, cosB, cosC, and λ are positive. Thus, the triangle is acute.

In addition, we have

4 =λ cosA

bc+λ cosB

ca+λ cosC

ab.

Since a cosA+b cosB+c cosC =2F

Rand 4RF = abc (where R is the circumradius

of the triangle), we readily find λ =a2b2c2

2F 2and obtain

x =a2(b2 + c2 − a2)

4F 2, y =

b2(c2 + a2 − b2)

4F 2, z =

c2(a2 + b2 − c2)

4F 2.

Note that conversely, if given an acute triangle, then these numbers x, y, z arepositive and satisfy the two equations: x + y + z = 4 is readily checked; also wehave b2z + c2y = c2x + a2z = b2x + a2y = λ, hence the calculations made at thebeginning (with equality in (1)) show that the second equation holds as well.

4006. Proposed by Dragolijub Milosevic.

Let x, y, z be positive real numbers such that xyz = 1. Prove that

2

xy + yz + zx− 1

x+ y + z≤ 1

3.

We received 15 correct solutions from 14 submitters. Ten of these solutions werealong the lines of the solution presented below, with variations in how they justifiedthe ancillary inequalities and how straightforwardly they handled the algebra. Inaddition, there was a MAPLE-based solution, which seemed heavy-handed for thisproblem. There were four other solutions that were defective in some way. Wepresent the solution by Henry Ricardo.


38/ SOLUTIONS

Let p = x+ y+ z, q = xy+ yz+ zx and r = xyz = 1. Observe that q2 ≥ 3rp = 3p,since, by the arithmetic-geometric means inequality,

q2 =1

2(x2y2 + y2z2) +

1

2(y2z2 + z2x2) +

1

2(z2x2 + x2y2) + 2xyz(x+ y + z)

≥ xy2z + yz2x+ zx2y + 2xyz(x+ y + z)

= 3xyz(x+ y + z) = 3rp = 3p.

The difference between the two sides of the inequality is one-third of

1− 6

q+

3

p≥ 1− 6

q+

9

q2=

Å1− 3

q

ã2≥ 0,

and the result follows with equality if and only if x = y = z = 1.

Editor’s comment. Oliver Geupel notes that this problem is equivalent to a prob-lem proposed by Vasile Cırtoaje and Mircea Lascu for the Junior TST 2003 Ro-mania. It is also Problem 72 in Chapter 20 of Inequalities, Theorems, Techniquesand Selected Problems by Zdravko Cvetkovski (Springer, 2012).

4007. Proposed by Mihaela Berindeanu.

Show that for any numbers a, b, c > 0 such that a2 + b2 + c2 = 12, we have

(a3 + 4a+ 8)(b3 + 4b+ 8)(c3 + 4c+ 8) ≤ 243.

We received nine submission of which eight were correct and complete. We presenttwo solutions.

Solution 1, by Angel Plaza.

By taking logarithms, the proposed inequality may be written as

ln(a3 + 4a+ 8

)+ ln

(b3 + 4b+ 8

)+ ln

(c3 + 4c+ 8

)3

≤ ln 24.

Changing variables a2 = x, b2 = y, c2 = z the problem becomes:

For any x, y, z > 0 such that x+ y + z = 12, prove that

ln(x3/2 + 4x1/2 + 8

)+ ln

(y3/2 + 4y1/2 + 8

)+ ln

(z3/2 + 4z1/2 + 8

)3

≤ ln 24.

Let us consider function f(x) = ln(x3/2 + 4x1/2 + 8

)for x > 0. Then

f ′′(x) =−8x3/2 − 3x5/2 + 12x− 16

√x− 16

2x3/2(x3/2 + 4

√x+ 8

)2and since f ′′(x) < 0 for x > 0, the function f is concave. By Jensen’s inequality

f(x) + f(y) + f(z)

3≤ f

(x+ y + z

3

)= f(12/3) = f(4) = ln 24.


SOLUTIONS /39

Solution 2, by the proposer.

Observe that (a− 2)4 ≥ 0 implies that a4 − 8a3 + 24a2 − 32a+ 16 ≥ 0, that is

a4 + 24a2 + 80 ≥ 8a3 + 32a+ 64,

which gives(a2 + 4)(a2 + 20) ≥ 8(a3 + 4a+ 8)

and hence

a3 + 4a+ 8 ≤ (a2 + 4)(a2 + 20)

8.

So,

(a3 + 4a+ 8)(b3 + 4b+ 8)(c3 + 4c+ 8)

≤ (a2 + 4)(a2 + 20)

8· (b2 + 4)(b2 + 20)

8· (c2 + 4)(c2 + 20)

8,

but we know that 3√xyz ≤ x+ y + z

3, therefore

(a2 + 4)(b2 + 4)(c2 + 4) ≤Åa2 + b2 + c2 + 12

3

ã3= 83

and

(a2 + 20)(b2 + 20)(c2 + 20) ≤Åa2 + b2 + c2 + 60

3

ã3= 243.

Finally,

(a3 + 4a+ 8

) (b3 + 4b+ 8

) (c3 + 4c+ 8

)≤ 83 · 243

83= 243.

Editor’s Comments. Angel Plaza sent two solutions: the second solution consistsin taking the logarithms of the given inequality, setting a2 = x, b2 = y, c2 = z,considering the concave function (on x ∈ (0, 12)) f(x) = ln(x3/2 + 4x1/2 + 8)and using Jensen’s Inequality. A very similar approach was also used by SefketArslanagic.

4008. Proposed by Mehmet Sahin.

Let ABC be a triangle with ∠ACB = 2α, ∠ABC = 3α, AD is an altitude andAE is a median such that ∠DAE = α. If |BC| = a, |CA| = b, |AB| = c, provethat

a

b= 1 +

…2(cb

)2− 1.

We received 15 correct solutions and one incorrect submission. We present thesolution given by Titu Zvonaru, modified slightly by the editor.


40/ SOLUTIONS

We have AD = c sin 3α,BD = c cos 3α, so DE =a

2− cos 3α. By the law of sines,

we havea

sin (180◦ − 5α)=

c

sin 2α, so a =

c sin 5α

sin 2α. Then

sinα

cosα= tanα =

DE

AD=

a2 − cos 3α

c sin 3α=

sin 5α− 2 sin 2α cos 3α

2 sin 2α sin 3α

=sin 5α− (sin 5α+ sin (−α))

2 sin 2α sin 3α

=sinα

2 sin 2α sin 3α,

so that

cosα = 2 sin 2α sin 3α = cosα− cos 5α,

which implies that

cos 5α = 0, so 5α = 90◦, or α = 18◦.

Hence, ∠BAC = 180◦ − 5α = 90◦, ∠ABC = 3α = 54◦ and ∠ACB = 2α = 36◦.

Since cos 36◦ =1 +√

5

4, we have

b = a cos 2α =

Ç1 +√

5

4

åa,

so

c =√a2 − b2 =

a2 − 3 +

√5

8a2 = a

5−√

5

8.

Now,a

b=

4

1 +√

5=√

5− 1 and

2(cb

)2− 1 = 2

Ç5−√

5

8

åÅ4

1 +√

5

ã2− 1 =

14− 6√

5

6 + 2√

5=

7− 3√

5

3 +√

5.


SOLUTIONS /41

Therefore, we have the following equivalences:

a

b= 1 +

…2(cb

)2− 1

⇐⇒√

5− 1 = 1 +

√7− 3

√5

3 +√

5

⇐⇒ (√

5− 2)2 =7− 3

√5

3 +√

5

⇐⇒ (9− 4√

5)(3 +√

5) = 7− 3√

5

⇐⇒ 7− 3√

5 = 7− 3√

5,

which is true and our proof is complete.


Let ma,mb,mc be the lengths of the medians of a triangle ABC. Prove that

1

ma+

1

mb+

1

mc≤ R

2r2,

where r and R are inradius and circumradius of ABC, respectively.

We received eleven solutions, of which ten were correct. We present two solutions.


Let F, s and ha, hb, hc be the area, semiperimeter, and altitudes of the triangle.Since mx ≥ hx, x ∈ {a, b, c} and

1

ha+

1

hb+

1

hc=

a

2F+

b

2F+

c

2F=

s

2F=

1

r

then1

ma+

1

mb+

1

mc≤ 1

ha+

1

hb+

1

hc=

1

r≤ R

2r2

because1

r≤ R

2r2⇐⇒ 2r ≤ R,

by Euler’s Inequality.

Solution 2, by Edmund Swylan.

We take it as known that the triangle with side lengths 2ma, 2mb, 2mc has mediansof lengths 3

2a, 32b,

32c. (See the drawing below.)


42/ SOLUTIONS

Let the area of 4ABC be F . The area of the big triangle is then 3F . Let thealtitudes of the big triangle be Ha, Hb, Hc.

We have that6F

2mx= Hx and Hx ≤

3

2x, for each x ∈ {a, b, c}. Therefore,

3F (1

ma+

1

mb+

1

mc) ≤ 3

2(a+ b+ c);

equality occurs if and only if the big triangle, and consequently 4ABC too, isequilateral. Finally,

3

2(a+ b+ c) = 3F

1

r≤ 3F

1

r

R

2r= 3F

R

2r2;

equality occurs if and only if 4ABC is equilateral.

4010. Proposed by Ovidiu Furdui.

Let f : [0, π2 ]→ R be a continuous function. Calculate

limn→∞

n

∫ π2

0

Åcosx− sinx

cosx+ sinx

ã2nf(x)dx.

There were eight submitted solutions for this problem, all of which were correct.We present two solutions.

Solution 1, by the group of M. Bello, M. Benito, O. Ciaurri, E. Fernandez, andL. Roncal, expanded slightly by the editor.

The value of the required limit is1

4

(f(0) + f

(π2

)). Indeed, if we denote by L

the limit, then from the identity

cos(x)− sin(x)

cos(x) + sin(x)= tan

(π4− x),


SOLUTIONS /43

we have

L = limn→∞

n

∫ π/2

0

(tan

(π4− x))2n

f(x) dx

= limn→∞

n

∫ π/4

−π/4(tan(s))2nf

(π4− s)ds

= limn→∞

n

∫ π/4

0

(tan(s))2n(f(π

4− s)

+ f(π

4+ s))

ds

= limn→∞

∫ 1

0

r1+1/n

1 + r2/n

(f(π

4− arctan(r1/n)

)+ f

(π4

+ arctan(r1/n)))

dr,

where we have used symmetry, and in the last step we have used the change ofvariable r = (tan(s))n.

Since f is a continuous function, ∃M such that |f(x)| ≤M , for x ∈ [0, π2 ], and∣∣∣∣∣ r1+1/n

1 + r2/n

(f(π

4− arctan(r1/n)

)+ f

(π4

+ arctan(r1/n)))∣∣∣∣∣ ≤M

for all r ∈ [0, 1], using the bound for f and that the fraction in r is bounded aboveby r/(1 + r2) (which is bounded by 1/2, by looking at (r − 1)2 ≥ 0). In this way,we can apply the dominated convergence theorem to obtain

L =

∫ 1

0

limn→∞

r1+1/n

1 + r2/n

(f(π

4− arctan(r1/n)

)+ f

(π4

+ arctan(r1/n)))

dr

=1

2

(f(0) + f

(π2

))∫ 1

0

r dr =1

4

(f(0) + f

(π2

)).


We show that the required limit is f(0)+f(π/2)4 . Let

In =

∫ π2

0

Åcosx− sinx

cosx+ sinx

ã2nf(x) dx =

∫ π2

0

(tan

(π4− x))2n

f(x) dx.

The change of variables x = π4 − tan−1(y) yields

In =

∫ 1

−1

y2n

1 + y2f(π

4− tan−1(y)

)dy.

But we have∫ 0

−1

y2n

1 + y2f(π

4− tan−1(y)

)dy =

∫ 1

0

u2n

1 + u2f(π

4+ tan−1(u)

)du

so that

In =

∫ 1

0

y2ng(y) dy,


44/ SOLUTIONS

where g(y) =(f(π4 − tan−1(y)

)+ f

(π4 + tan−1(y)

))· 1y2+1 . It is known that if g

is continuous on [0, 1], then limn→∞

n∫ 1

0xng(x) dx = g(1) [for completeness, a quick

proof is given at the end]. From this result, it follows that

limn→∞

(2n) · In = g(1) =f(0) + f(π/2)

2

and so

limn→∞

n · In =f(0) + f(π/2)

4,

as claimed.

For the proof of the property used above, let ε > 0. Using the continuity of g, wechoose δ ∈ (0, 1) such that |g(x)− g(1)| ≤ ε whenever x ∈ [δ, 1]. Then we have∣∣∣∣∣(n+ 1)

∫ 1

0

xn · g(x) dx− g(1)

∣∣∣∣∣ =

∣∣∣∣∣(n+ 1)

∫ 1

0

xn · (g(x)− g(1)) dx

∣∣∣∣∣≤ (n+ 1)

∫ δ

0

xn|g(x)− g(1)| dx+ (n+ 1)

∫ 1

δ

xn|g(x)− g(1)| dx

≤M · δn+1 + ε

where M denotes the maximum of the continuous function x 7→ |g(x) − g(1)| on

[0, 1]. Since 0 < δ < 1, we deduce lim supn→∞

|(n+ 1)∫ 1

0xn · g(x) dx− g(1)| ≤ ε. Since

the latter holds for any positive ε, we must have

limn→∞

Ç(n+ 1)

∫ 1

0

xn · g(x) dx− g(1)

å= 0,

and the result follows.

Editor’s Comments. This type of problem has its roots in Fourier analysis, wherewe are interested in limits such as the one in the problem statement. This particu-lar limit picks out half the arithmetic mean of the function’s value at the endpointsof the interval [0, π2 ]; more classical Fourier analysis will focus on limits which pickout the function’s value at a specific point, like the ‘Dirac delta’ distribution. Allsolutions aside from Solution 1 essentially followed Bataille’s Solution 2, includingthe proposer’s; A. Plaza’s solution used a limit result from the proposer’s ownbook (O. Furdui, Limits, Series and Fractional Part Integrals, Springer, Seconded., 2013) to skip a substitution. The techniques in both solutions (i.e. utiliz-ing dominated convergence in a clever way, and separating an integral up into twoparts which are handled using the two different functions involved in the integrand)are common techniques in classical analysis. A. Stadler’s approach (namely, using‘Big O’ notation instead of more precise estimates) is equally successful and iscommon in analytic number theory.


SOLUTIONS /79



4011. Proposed by Abdilkadir Altinas.

In non-equilateral triangle ABC, let H be the orthocentre of ABC and J be theorthocentre of the orthic triangle DEF of ABC (that is the triangle formed bythe feet of the altitudes of ABC). If ∠BAC = 60◦, show that AJ ⊥ HJ .

We received nine solutions. Eight of the solutions used angle-chasing in cyclicquadrilaterals, and one solution used barycentric coordinates. The former type ofsolutions were simpler, but they all missed the fact that there are subtleties if theorthocenter is not interior to the triangle.

We present the solution by Ricardo Barroso Campos slightly modified by the editor.

Since ∠HFA = ∠HEA = 90◦, quadrilateral AFHE is cyclic. Hence

∠EHF = 180◦ − ∠CAB = 120◦.

Denote by Γ the circumcircle of AFHE, and note that AH is the diameter of thiscircle (since ∠AFH = 90◦).

In the cyclic quadrilateral BDHF , ∠BDF = ∠BHF = 180◦ − ∠EHF = 60◦.Similarly, from the cyclic quadrilateral CDHE, we get ∠CDE = 60◦. Hence∠FDE = 180◦ − (∠BDF + ∠CDE) = 60◦.


80/ SOLUTIONS

Denote the feet of the altitudes from D, E, and F by K, L and M respectively,as in the diagram. DLJM is cyclic, thus ∠LJM = 180◦ −∠FDE = 120◦. Hence∠FJE = ∠LJM = 120◦, which implies that the quadrilateral AFJE is cyclic(since ∠FJE + ∠FAE = 180◦). It follows that J is on the circle Γ. Hence, sinceAH is the diameter of Γ, we get ∠AJH = 90◦, so AJ ⊥ HJ .

Editor’s Comments. The provided solution fails if one of the orthocentres is notinterior to its triangle. The following diagram shows the case where the point Jis not interior to DEF (for one, ∠FJE = 60◦, not 120◦). Note however that it isnot difficult to adjust the solution for these cases.

4012. Proposed by Leonard Giugiuc.

Let n be an integer with n ≥ 3. Consider real numbers ak, 1 ≤ k ≤ n such that

a1 ≥ a2 ≥ . . . ≥ an−1 ≥ 1 ≥ an ≥ 0 andn∑k=1

ak = n.

Prove that(n− 2)(n+ 1)

2≤

∑1≤i<j≤n

aiaj ≤n(n− 1)

2.

We received eight submissions of which seven were correct and complete. Wepresent the solution by Ivan Chan Kai Chin.

Since we have

2 ·∑

1≤i≤j≤naiaj = (

n∑k=1

ak)2 −n∑k=1

a2k = n2 −n∑k=1

a2k,

Crux Mathematicorum, Vol. 42(2), February 2016

SOLUTIONS /81

it suffices to prove that

n ≤n∑k=1

a2k ≤ n+ 2.

The left inequality holds by Cauchy-Schwarz, since

n∑k=1

a2k ≥1

n· (

n∑k=1

ak)2 = n,

with equality when a1 = a2 = · · · = an = 1.

For the other inequality, set bk = ak − 1 for all 1 ≤ k ≤ n. Then bk ≥ 0 for all

1 ≤ k ≤ n− 1, −1 ≤ bn ≤ 0, andn∑k=1

bi = 0. We have

n∑k=1

a2k =n∑k=1

(1 + bk)2 =n−1∑k=1

(1 + bk)2 + (1− (b1 + b2 + · · ·+ bn−1))2. (1)

Define the quantity S = b1 + b2 + · · ·+ bn−1 ≤ 1, and

f(b1, b2, · · · , bn−1) =n−1∑k=1

(1 + bk)2.

For any 1 ≤ i ≤ j ≤ n− 1,

f(· · · bi, · · · , bj , · · · ) ≤ f(· · · bi + bj , · · · , 0, · · · ),

since(1 + bi)

2 + (1 + bj)2 ≤ (1 + bi + bj)

2 + 1 ⇐⇒ 2bibj ≥ 0

holds for all bi, bj , 1 ≤ i ≤ j ≤ n− 1. Thus we have

f(b1, b2, · · · , bn−1) ≤ f(b1 + b2 + · · · bn−1, 0, · · · , 0) = f(S, 0, · · · , 0)

and (1) becomes

f(S, 0, · · · , 0) + (1− S)2 = (1 + S)2 + (n− 2) + (1− S)2

= n+ 2S2

≤ n+ 2

Equality holds when S = 1, b2 = b3 = · · · = bn−1 = 0, b1 = 1 and bn = −1, whichcorresponds to a1 = 2, a2 = a3 = · · · = an−1 = 1, an = 0.

4013. Proposed by Mehmet Sahin.

Let a, b, c be the sides of triangle ABC, D be the foot of the altitude from A andE be the midpoint of BC. Define θ = ∠DAE and suppose that ∠ACB = 2θ.Prove that the sides of the triangle satisfy

(a− b)2 = 2c2 − b2.


82/ SOLUTIONS

We received 16 submissions. Among them one simply stated that the claim wasincorrect and provided a counterexample, while 15 proved the claim under the ad-ditional assumption that b > c; moreover, 5 proved that for the claim to be correct,the assumption that b > c is both necessary and sufficient, and 3 of those submis-sions went on to provide a complete description of triangles that satisfy the givenhypotheses.

We present the solution by Joel Schlosberg, supplemented by ideas from C. R. Prane-sachar.

We shall prove that if a triangle satisfies ∠ACB = 2∠DAE and, moreover, b > c,then (a − b)2 = 2c2 − b2; if b < c then a = 2b (and the claimed equation fails tohold). Note that if b = c then A = D = E is the midpoint of the segment BC,and the triangle is degenerate.

Scale 4ABC so that b = 1. By right-angle trigonometry, AD = sin 2θ =2 sin θ cos θ, so that

AD2 = 4 sin2 θ(1− sin2 θ).

Use signed lengths for segments on BC, with BC positive. Then

DE = pAD tan θ = 2p sin2 θ,

where p = 1 if B and D are on one side of E, and C is on the other (which happensif and only if b > c); otherwise, when E is between B and D (and, equivalently,b < c) then we set p = −1. Furthermore, we have

DC = cos 2θ = 1− 2 sin2 θ

EC = DC −DE = 1− (2 + 2p) sin2 θ

a = BC = 2EC = 2− (4 + 4p) sin2 θ

BD = BC −DC = 1− (2 + 4p) sin2 θ.

By the Pythagorean theorem (using p2 = 1),

c2 = AD2 +BD2 = 1− 8p sin2 θ + (16 + 16p) sin4 θ, (1)

while (using b = 1)

(a− b)2 + b2

2=

1

2a2 − ab+ b2 = 1− (4 + 4p) sin2 θ + (16 + 16p) sin4 θ. (2)

Comparing equations (1) and (2), we see that (a − b)2 = 2c2 − b2 iff p = 1. Onthe other hand, setting p = −1 and b = 1 in equation (2) we get a = 2 and deducethat a = 2b.

Editor’s Comments. This problem should be compared with problem 4008 whosesolution appeared in the previous issue. It dealt with triangles for which ∠ACB =2∠DAE and, in addition, ∠ABC = 3∠DAE. One finds that this can happen if


SOLUTIONS /83

and only if ∠A = 90◦,∠B = 54◦, and ∠C = 36◦; of course this implies that b > cand, consequently, that (a− b)2 = 2c2 − b2.

4014. Proposed by Mihaela Berinedanu.

Let n be a natural number and let x, y and z be positive real numbers such thatx+ y + z + nxyz = n+ 3. Prove that

(1 +y

x+ nyz)(1 +

z

y+ nzx)(1 +

x

z+ nxy) ≥ (n+ 2)3

and determine when equality holds.

We received six correct solutions. We present the solution by Dionne Bailey, ElsieCampbell and Charles Diminnie (joint).

The arithmetic-geometric means inequality yields that

n+3 = x+y+z+nxyz ≥ (n+3) [x · y · z · (xyz)n]1/n+3

= (n+3) [xyz](n+1)/(n+3)

,

so that xyz ≤ 1.

The inequality is equivalent to

(x+ y + nxyz)(y + z + nxyz)(z + x+ nxyz) ≥ (n+ 2)3(xyz)

or(n+ 3− z)(n+ 3− x)(n+ 3− y) ≥ (n+ 2)3xyz.

Using the arithmetic-geometric means inequality and the fact that xyz ≤ (xyz)2/3,we obtain that

(n+ 3− x)(n+ 3− y)(n+ 3− z)= (n+ 3)3 − (n+ 3)2(x+ y + z) + (n+ 3)(xy + yz + zx)− xyz≥ (n+ 3)3 − (n+ 3)2[(n+ 3)− nxyz] + 3(n+ 3)(xyz)2/3 − xyz≥ n(n+ 3)2xyz + 3(n+ 3)xyz − xyz= (n3 + 6n2 + 9n+ 3n+ 9− 1)xyz = (n+ 2)3xyz,

as desired, with equality if and only if x = y = z = 1.


Find all real numbers a such that

a cosx+ (1− a) cosx

3>

sinx

x

for every nonzero x of the interval (− 3π2 ,

3π2 ).

There were four submitted solutions for this problem, all of which were correct.We present the solution by Joel Schlosberg.


84/ SOLUTIONS

We will prove that the inequality

a cos(x) + (1− a) cos(x

3

)>

sin(x)

x(1)

holds for x ∈ (− 3π2 ,

3π2 ) \ {0} if and only if a ∈ (−∞, 1/4].

Substituting y = x/3 and dividing by y2, the above inequality is equivalent to

a

Åcos(y)− cos(3y)

y2

ã<

3y cos(y)− sin(3y)

3y3, (2)

for |y| ∈ (0, π/2). By repeated applications of l’Hospital’s Rule,

limy→0

cos(y)− cos(3y)

y2= limy→0

− sin(y) + 3 sin(3y)

2y= limy→0

− cos(y) + 9 cos(3y)

2= 4,

and similarly,

limy→0


3y3= limy→0

3 cos(y)− 3y sin(y)− 3 cos(3y)

9y2

= limy→0

−6 sin(y)− 3y cos(y) + 9 sin(3y)

18y

= limy→0

−9 cos(y) + 3y sin(y) + 27 cos(3y)

18= 1.

Suppose that (1) holds for x ∈ (− 3π2 ,

3π2 ) \ {0}. Then the rewritten inequality (2)

holds for non-zero y in a neighbourhood of 0. Taking y → 0 on both sides yields4a ≤ 1, and so a ∈ (−∞, 1/4].

Since both sides of (2) are even functions of y, it is sufficient to prove (2) fory ∈ (0, π/2). We let

f(x) = 9y cos(y)− 4 sin(3y) + 3y cos(3y).

For y ∈ (0, π/2), it is well-known that tan(y) > y, so

f ′(y) = 9(cos(y)− y sin(y)− cos(3y)− y sin(3y))

= 9(4(1− cos2(y)) cos(y)− 4y(1− sin2(y)) sin(y))

= 36(sin2(y) cos(y)− y cos2(y) sin(y))

= 36 cos2(y) sin(y)(tan(y)− y) > 0

(via triple-angle formulas). Therefore, for y ∈ (0, π/2), f(y) > f(0) = 0, which isequivalent to (2) when a = 1/4. Hence (1) holds for a = 1/4.

Suppose now that a < 1/4. For y ∈ (0, π/2), we have

cos(y)− cos(3y) = 4(1− cos2(y)) cos(y) = 4 sin2(y) cos(y) > 0,


SOLUTIONS /85

so that

a

Åcos(y)− cos(3y)

y2

ã<

1

4

Åcos(y)− cos(3y)

y2

ã<


3y3,

and we are done.

Editor’s Comments. All four solution methods involved similar elements: trigono-metric identities used to rewrite the inequality, a limit (either by power series orby L’Hospital’s Rule), and some calculus. Deiermann noted that if we set theright-hand side of the original inequality equal to 1 for x = 0, then we may allowequality at x = 0. Deiermann also suggested a generalization, propped up byMathematica: if n ≥ 3, then we have

a cos(x) + (1− a) cos(xn

)>

sin(x)

x.

for all non-zero x ∈ (− 3π2 ,

3π2 ) \ {0} if and only if a ≤ n2−3

3(n2−1) . A quick sketch of

the argument by the editor seems to indicate that it is true, but the conclusion ofthe proof is still out of reach.


Let x, y, z be positive real numbers. Find the maximal value of the expression

x+ 2y

2x+ 3y + z+

y + 2z

2y + 3z + x+

z + 2x

2z + 3x+ y.

We received 21 submissions, all of which were correct. We present two solutions.


Let S (x, y, z) denote the given expression. Then by using Cauchy-Schwarz In-equality we have

3− S (x, y, z) =∑cyc

Å1− x+ 2y

2x+ 3y + z

ã=∑cyc

x+ y + z

2x+ 3y + z

=6 (x+ y + z)

6

∑cyc

1

2x+ 3y + z

=1

6

∑cyc

(2x+ 3y + z) ·∑cyc

1

2x+ 3y + z

≥ 1

6· 9 =

3

2.

Hence, S (x, y, z) ≤ 3

2and S (x, x, x) =

3

2.


86/ SOLUTIONS

Solution 2, by Sefket Arslanagic.

Since the given inequality is homogeneous, we may assume that x+ y+ z = 1. Bythe AM-HM Inequality, we have

S(x, y, z) = 3−Å

1

1 + x+ 2y+

1

1 + y + 2z+

1

1 + z + 2x

ã≤ 3− 9

(1 + x+ 2y) + (1 + y + 2z) + (1 + z + 2x)

= 3− 9

6=

3

2.

Hence, the maximum value of S(x, y, z) is 32 attained when x = y = z.

Editor’s Comments. Kee-Wai Lau made an interesting and not-so-easy-to-seeobservation that

S(x, y, z)− 3

2= −

∑(3x+ y + 2z)(x+ y − 2z)2

6∏

(2x+ 3y + z)≤ 0.


Let P be a point of the incircle γ of a triangle ABC. The perpendiculars toBC,CA and AB through P meet γ again at U, V and W , respectively. Prove thatone of the numbers PU ·BC,PV · CA,PW ·AB is the sum of the other two.

From the 6 correct submissions we received, we present a composite of the similarsolutions by Sefket Arslanagic, Ricard Peiro i Estruch, and Joel Schlosberg.

Since PV ⊥ CA and PW ⊥ AB, ∠V PW is either equal to or supplementary to∠BAC, so

sin∠V PW = sin∠BAC = sinA;

similarly,

sin∠WPU = sinB and sin∠UPV = sinC.

Moreover, because PUVW is cyclic we have

sin∠V PW = sin∠V UW, sin∠WPU = sin∠WV U, sin∠UPV = sin∠UWV.

Finally, the Law of Sines applied to ∆UVW implies

VW

WU=

sin∠V UWsin∠WV U

andUV

WU=

sin∠UWV

sin∠WV U,

while applied to ∆ABC implies

sinA

sinB=BC

CAand

sinC

sinB=AB

CA.


SOLUTIONS /87

Let us suppose that the diagram has been labeled so that the quadrilateral PUVWis cyclic in that order, whence PV is the diagonal, and Ptolemy’s theorem saysthat PV ·WU = PU · VW + PW · UV . Putting it all together, we get

PV = PU · VWWU

+ PW · UVWU

= PU · sin∠V UWsin∠WV U

+ PW · sin∠UWV

sin∠WV U

= PU · sin∠V PWsin∠WPU

+ PW · sin∠UPVsin∠WPU

= PU · sinA

sinB+ PW · sinC

sinB

= PU · BCCA

+ PW · ABCA

.

Thus, we conclude that PV · CA = PU · BC + PW · AB; in other words, theproduct involving the diagonal of the quadrilateral equals the sum of the productsinvolving the sides.

Additionally, the proposer observed (and proved) that the area of ∆UVW is in-dependent of the choice of P on γ.


Let

In =

∫ 1

0

· · ·∫ 1

0

ln(x1x2 · · ·xn) ln(1− x1x2 · · ·xn)dx1dx2 · · · dxn,

where n ≥ 1 is an integer. Prove that this integral converges and find its value.

We received three solutions, all of which were correct and complete. We presentthe solution by the proposer.

The integral equals

n (n+ 1− ζ(2)− ζ(3)− · · · − ζ(n+ 1)) ,

where ζ denotes the Riemann zeta function.

We have, based on symmetry reasons, that for all i, j = 1, 2, . . . , n

∫ 1

0

· · ·∫ 1

0

lnxi ln(1− x1x2 · · ·xn)dx1dx2 · · · dxn

=

∫ 1

0

· · ·∫ 1

0

lnxj ln(1− x1x2 · · ·xn)dx1dx2 · · · dxn,


88/ SOLUTIONS

and this implies that

In = n

∫ 1

0

· · ·∫ 1

0

lnx1 ln(1− x1x2 · · ·xn)dx1dx2 · · · dxn

= n

∫ 1

0

· · ·∫ 1

0

− lnx1

∞∑k=1

(x1 · · ·xn)k

kdx1dx2 · · · dxn

(∗)= n

∞∑k=1

1

k

∫ 1

0

· · ·∫ 1

0

− lnx1(x1 · · ·xn)kdx1dx2 · · · dxn

= n∞∑k=1

1

k

∫ 1

0

(−xk1 lnx1)dx1

∫ 1

0

xk2dx2 · · ·∫ 1

0

xkndxn

= n∞∑k=1

1

k(k + 1)n+1.

We used at step (*) Tonelli’s Theorem for nonnegative functions, which allows usto interchange the integration sign and the summation sign.

Let Sn+1 =∞∑k=1

1

k(k + 1)n+1. Since

1

k(k + 1)n+1=

1

k(k + 1)n− 1

(k + 1)n+1,

we have, by summation, that Sn+1 = Sn − (ζ(n + 1) − 1). This implies, sinceS1 = 1, that

Sn+1 = S1− (ζ(2)+ ζ(3)+ · · ·+ ζ(n+1)−n) = n+1− ζ(2)− ζ(3)−· · ·− ζ(n+1).

Hence

In = n (n+ 1− ζ(2)− ζ(3)− · · · − ζ(n+ 1)) ,

and the problem is solved.


A triangle with side lengths a, b, c has perimeter 3. Prove that

a3 + b3 + c3 + a4 + b4 + c4 ≥ 2(a2b2 + b2c2 + c2a2).

We received 21 correct solutions. We present the solution by AN-anduud ProblemSolving Group.

The claimed inequality is equivalent to

(a3 + b3 + c3)(a+ b+ c) + 3(a4 + b4 + c4) ≥ 6(a2b2 + b2c2 + c2a2)


SOLUTIONS /89

or

[(a3b+ b3a) + (a3c+ c3a) + (b3c+ c3b)] + [2(a4 + b4) + 2(b4 + c4) + 2(c4 + a4)]

≥ 6(a2b2 + b2c2 + c2a2).

By the AM-GM Inequality we have

a3b+ b3a ≥ 2a2b2, a3c+ c3a ≥ 2a2c2, b3c+ c3b ≥ 2b2c2

anda4 + b4 ≥ 2a2b2, b4 + c4 ≥ 2b2c2, c4 + a4 ≥ 2a2c2.

Adding the above inequalities, we obtain the desired inequality. Equality holds ifand only if a = b = c = 1.

4020. Proposed by Leonard Giugiuc and Daniel Sitaru.

Let ABC be a triangle and let the internal bisectors from A,B and C intersectthe sides BC,CA and AB in D,E and F , respectively. The incircle of ∆ABCtouches the sides BC,CA and AB in M,N , and P , respectively. Prove that[MNP ] ≤ [DEF ], where [·] denotes the area of the specified triangle.

We received eleven submissions, of which nine were correct and complete. Wepresent the solution by Sefket Arslanagic, slightly modified by the editor.

Denote by α, β and γ the angles BAC, ABC and respectively ACB of the triangle,and let r be the radius of the incircle.

From the quadrilateral PIMB note that ∠PIM = 180◦ − ∠ABC = 180◦ − β,whence

[PIM ] =PI ·MI

2sin(∠PIM) =

r2

2sin(180◦ − β) =

r2

2sinβ.

Similarly, we calculate [MIN ] and [NIP ], and we get

[MNP ] = [PIM ] + [MIN ] + [NIP ]

=r2

2· (sinβ + sin γ + sinα).

(1)


90/ SOLUTIONS

On the other hand, we have ∠FID = ∠AIC = 180◦ − α2 −

γ2 , and so

[FID] =ID · IF

2sin(∠FID) =

ID · IF2

sinα+ γ

2.

Similarly, calculate the area of 4EIF and 4DIE. We have

[DEF ] = [DIE] + [EIF ] + [FID]

=ID · IE

2sin

α+ β

2+IE · EF

2sin

β + γ

2+ID · IF

2sin

α+ γ

2.

(2)

The triangles4PIF ,4MID and4NIE are all right-angled triangles, from whichit follows that IF ≥ r, ID ≥ r and IE ≥ r. Hence, from the formula for [DEF ]above we get

[DEF ] ≥ r2

2

Åsin

α+ β

2+ sin

β + γ

2+ sin

α+ γ

2

ã.

Comparing this with the formula for [MNP ] in (1), in order to show that [MNP ] ≤[DEF ] it is sufficient to show that

sinα+ sinβ + sin γ ≤ sinα+ β

2+ sin

β + γ

2+ sin

α+ γ

2. (3)

However, using the sum to product trigonometric formula, we have

sinα+ sinβ = 2 sinα+ β

2cos

α− β2≤ 2 sin

α+ β

2,

where the inequality follows from the fact that sin α+β2 ≥ 0 and cos α−β2 ≤ 1.

Similarly we have sinβ + sin γ ≤ 2 sin β+γ2 and sin γ + sinα ≤ 2 sin α+γ

2 , which wecan add to get the inequality in (3), the final step we need in order to concludethat [MNP ] ≤ [DEF ].


126/ SOLUTIONS



4021. Proposed by Arkady Alt.

Let (an)n≥0 be a sequence of Fibonacci vectors defined recursively by a0 = a, a1 =

b and an+1 = an + an−1 for all integers n ≥ 1. Prove that, for all integers n ≥ 1,the sum of vectors a0 + a1 + · · ·+ a4n+1 equals kai for some i and constant k.

We received nine correct solutions. We present the solution by David Stone andJohn Hawkins (joint).

We shall prove that a0 + a1 + · · · + a4n+1 = L2n+1a2n+2, where Lk is the kthLucas number. We use some easily proven results. Here, Fk is the kth Fibonaccinumber.

1. F0 + F1 + · · ·+ Fm = Fm+2 − 1.

2. F4n+2 = L2n+1F2n+1

3. F4n+3 = L2n+1F2n+2 + 1

4. ak = Fk−1a0 + Fka1 for k ≥ 1.

Therefore,

m∑k=0

ak = a0 +m∑k=1

(Fk−1a0 + Fka1)

= a0 +

(m∑k=1

Fk−1

)a0 +

(m∑k=1

)a1

= a0 + (Fm+1 − 1)a0 + (Fm+2 − 1)a1

= Fm+1a0 + Fm+2a1 − a1= am+2 − a1

Hence, with m = 4n+ 1,

4n+1∑k=0

ak = a4n+3 − a1 = F4n+2a0 + F4n+3a1 − a1

= (L2n+1F2n+1)a0 + (L2n+1F2n+2)a1

= L2n+1(F2n+1a0 + F2n+2a1)

= L2n+1a2n+2.

Crux Mathematicorum, Vol. 42(3), March 2016

SOLUTIONS /127

Editor’s Comments. Various solvers expressed the coefficient of a2n+2 as L2n+1,

F2n + F2n+2, and F4n+2

F2n+1and variations of these resulting from different indexing

of the Fibonacci sequence. Swylan pointed out that if the word ‘constant’ isinterpreted to mean ‘independent of n’, then the claim of the problem is false.Perhaps ‘scalar’ would have been a better word.


In a triangle ABC, let internal angle bisectors from angles A,B and C intersectthe sides BC,CA and AB in points D,E and F and let the incircle of ∆ABCtouch the sides in M,N , and P , respectively. Show that

PA

PB+MB

MC+NC

NA≥ FA

FB+DB

DC+EC

EA.

We received eleven submissions, of which seven were correct, two were incorrect,and two were incomplete. We present the solution by Titu Zvonaru.

Define x = NA = PA, y = PB = MB, and z = MC = NC; then

BC = y + z, CA = z + x, and AB = x+ y.

By the angle bisector theorem we have

FA

FB=CA

BC=z + x

y + z,

DB

DC=AB

CA=x+ y

z + x, and

EC

EA=BC

AB=y + z

x+ y.

We therefore have to prove that for positive real numbers x, y, z,

x

y+y

z+z

x≥ z + x

y + z+x+ y

z + x+y + z

x+ y. (1)

After clearing denominators what we must prove reduces to

x2y4 +y2z4 +z2x4 +x3y3 +y3z3 +z3x3 ≥ x3yz2 +x2y3z+xy2z3 +3x2y2z2. (2)

By the AM-GM inequality we have

x2y4 + y2z4 + z2x4 ≥ 3x2y2z2,

x3y3 + z3x3 + z3x3 ≥ 3x3yz2,

y3z3 + x3y3 + x3y3 ≥ 3x2y3z, and

z3x3 + y3z3 + y3z3 ≥ 3xy2z3 ,

which together imply that (2) holds. Equality holds if and only if x = y = z,which immediately implies that the triangle is equilateral.

Editor’s Comments. Most submissions reduced our problem to equation (1), butthen algebra caused difficulties with two of the faulty arguments. The solutionfrom Salem Malikic neatly avoided calculations by remarking that (1) is known;


128/ SOLUTIONS

see, for example, the Belarussian IMO Team preparation tests of 1997, where cal-culations are much simplified by exploiting the cyclic symmetry of the inequalities.Beware, however, that one must not assume noncyclic symmetry (as in one of theincomplete submissions).

4023. Proposed by Ali Behrouz.

Find all functions f : R+ 7→ R+ such that for all x, y ∈ R with x > y, we have

f

Åx

x− y

ã+ f(xf(y)) = f(xf(x)).

We received two correct submissions. We present the solution by Joseph Ling.

It is easy to see that f (x) =1

x∀x > 0 satisfies

f

Åx

x− y

ã+ f (xf (y)) = f (xf (x)) (1)

whenever 0 < y < x. We show that there are no other solutions f : (0,∞)→ (0,∞)to (1).

First, we note that f is one-to-one. For if 0 < y < x are such that f (y) = f (x) ,

then (1) implies that fÄ

xx−y

ä= 0, which is impossible.

Second, we note that f (x) ≤ 1x for all x > 0. For if there exists x > 0 such that

f (x) > 1x , then y = x− 1

f(x) satisfies 0 < y < x. But then xx−y = xf (x) and (1)

will imply that f (xf (y)) = 0, which is impossible.

Now, suppose that 0 < y1 < y2. Consider x = y2 + 1f(y1)

. Then 0 < y1 < y2 < x

and (1) implies that

f

Åx

x− y2

ã+ f (xf (y2)) = f (xf (x)) = f

Åx

x− y1

ã+ f (xf (y1)) . (2)

By the definition of x, xx−y2 = xf (y1) . So, (2) is reduced to f (xf (y2)) =

fÄ

xx−y1

ä. Since f is one-to-one, we have xf (y2) = x

x−y1 , and so, x = y1 + 1f(y2)

.

Using this and the definition of x, we see that 1f(y1)

−y1 = 1f(y2)

−y2. Since y1 and

y2 are arbitrary, 1f(y) − y is independent of y, and so, it must be some constant,

say, c. That is,

f (y) =1

y + c

for all y > 0.

It remains to show that c = 0. Since f (x) ≤ 1x for all x, c ≥ 0. Furthermore, if


SOLUTIONS /129

c > 0, then for any 0 < y < x, we have

f

Åx

x− y

ã+ f (xf (y)) =

x− yx+ c (x− y)

+y + c

x+ c (y + c)

>x− y

x+ c (x+ c)+

y + c

x+ c (x+ c)=

x+ c

x+ c (x+ c)

= f (xf (x)) ,

a contradiction to (1). So, c = 0 and our proof is complete.


Let a, b, c and d be real numbers such that a2 + b2 + c2 + d2 = 4. Prove that

abc+ abd+ acd+ bcd+ 4 ≥ a+ b+ c+ d

and determine when equality holds.

We received three correct solutions. We present the solution by Titu Zvononu,modified by the editor.

We consider several cases separately.

Case 1. If a, b, c, d ≥ 0, then by the Cauchy-Schwarz Inequality, we have

(a+ b+ c+ d)2 ≤ (12 + 12 + 12 + 12)(a2 + b2 + c2 + d2) = 16.

Thus, a+ b+ c+ d ≤ 4 from which the result follows immediately.

Case 2. If a, b, c, d ≤ 0, we set x = −a, y = −b, z = −c, and t = −d. Then,x, y, z, t ≥ 0 and we would now like to show that

−(xyz + xyt+ xzt+ yzt) + 4 ≥ −(x+ y + z + t) or

2(x+ y + z + t)− (xyz + xyt+ xzt+ yzt) + 4 ≥ x+ y + z + t. (1)

Since we know x2 + y2 + z2 + t2 = 4, we have by the first case (with a, b, c, dreplaced with x, y, z, t, respectively and symbolically), that

x+ y + z + t ≤ 4. (2)

Then,

4(x+ y + z + t)− 2(xyz + xyt+ xzt+ yzt)

= (x2 + y2 + z2 + t2)(x+ y + z + t)− 2(xyz + xyt+ xzt+ yzt)

= x(y − z)2 + y(x− t)2 + z(x− t)2 + t(y − z)2 + x(x2 + t2)

+ y(y2 + z2) + z(z2 + y2) + t(t2 + x2)

≥ 0,

which together with (2) implies (1).


130/ SOLUTIONS

Case 3. If one of a, b, c, d is nonnegative and the other three are nonpositive.Due to the symmetry in the given equation and the one we wish to prove, wemay assume that a ≥ 0 and b, c, d ≤ 0. Here we let y = −b, z = −c, andt = −d. Then, a, y, z, t ≥ 0 with a2 + y2 + z2 + t2 = 4 and we wish to proveayz + ayt+ azt− yzt+ 4 ≥ a− y − z − t or

2(y + z + t) + ayz + ayt+ azt− yzt+ 4 ≥ a+ y + z + t. (3)

Now,

4(y + z + t) + 2(ayz + ayt+ azt− yzt)= (a2 + y2 + z2 + t2)(y + z + t) + 2(ayz + ayt+ azt− yzt)≥ y(z2 + t2)− 2yzt

= y(z − t)2

≥ 0,

so, 2(y + z + t) + ayz + ayt+ azt− yzt+ 4 ≥ 4, thus, establishing (3), as desired,since a+ y + z + t ≤ 4 (see Case 1).

Case 4. If a, b ≥ 0 and c, d ≤ 0, we set z = −c and t = −d. Then a, b, z, t ≥ 0 suchthat a2 + b2 + z2 + t2 = 4 and we would like to show that

−abz − abt+ azt+ bzt+ 4 ≥ a+ b− z − t or

2(z + t)− abz − abt+ azt+ bzt+ 4 ≥ a+ b+ z + t. (4)

Now,

4(z + t)− 2(abz + abt− azt− bzt)= (a2 + b2 + z2 + t2)(z + t)− 2(abz + abt− azt− bzt)= z(a− b)2 + t(a− b)2 + (z + t)(z2 + t2) + 2azt+ 2bzt

≥ 0.

So, 2(z+ t)−abz−abt+azt+ bzt ≥ 4, thus establishing (4) since a+ b+z+ t ≤ 4.

Case 5. If a, b, c ≥ 0 and d ≤ 0, we set t = −d. Then a, b, c, t ≥ 0 with a2 + b2 +c2 + t2 = 4 and we would like to show that

abc− abt− act− bct+ 4 ≥ a+ b+ c− t or

2t+ abc− abt− act− bct+ 4 ≥ a+ b+ c+ t. (5)

Since a+ b+ c+ t ≤ 4, to establish (5), it suffices to show that

(a2 + b2 + c2 + t2)t+ 2(abc− abt− act− bct) ≥ 0. (6)


SOLUTIONS /131

We let L denote the left-hand side of (6) and assume, without loss of generality,that a ≥ b ≥ c. Note that

L = t(b− c)2 + t(a− t)2 + 2a(t− b)(t− c) (7)

and L = t(a− b)2 + t(c− t)2 + 2c(t− a)(t− b). (8)

If t ≤ b, then from (7) we can see that L ≥ 0 and if t ≥ b, then L ≥ 0 from (8).Hence, we can conclude that (6) is true, as desired.

Examining the five cases, it is readily seen that equality can only hold in Case 5when a = b = c = t; that is, if and only if (a, b, c, d) = (1, 1, 1,−1) and all itspermutations.

4025. Proposed by Dragoljub Milosevic.

Prove that for positive numbers a, b and c, we have

3

Åa

2b+ c

ã2+

3

Åb

2c+ a

ã2+

3

Åc

2a+ b

ã2≥ 3√

3.

We received eleven correct solutions. We present the solution by Salem Madikic.

Let f(a, b, c) denote the left hand side of the given inequality. By the AM-GMInequality, we have

3

Åa

2b+ c

ã2=

a3√a(2b+ c)2

=a

3√

9 3

»a · 2b+c3 · 2b+c3

≥ 3a3√

9(a+ 2b+c

3 + 2b+c3

)=

3 3√

3a

3a+ 4b+ 2c.

Using similar inequalities involving the other two summands, we then have

f(a, b, c) ≥ 33√

3∑cyc

a

3a+ 4b+ 2c. (1)

Now, by the Cauchy-Schwarz Inequality, we haveÇ∑Å…a

3a+ 4b+ 2c

ã2åÅ∑(»a(3a+ 4b+ 2c)

)2ã≥∑

a2.

So, ∑ a

3a+ 4b+ 2c≥

∑a2∑

a(3a+ 4b+ 2c)=

∑a2

3∑a2 + 6

∑ab

=1

3. (2)

Substituting (2) into (1), f(a, b, c) ≥ 3√

3 follows immediately.


132/ SOLUTIONS

To achieve equality, we must have 3a = 2b+c, 3b = 2c+a, and 3c = 2a+b. Withoutloss of generality, we may assume that max{a, b, c} = a. Then, 3a = 2b+ c implies2(a − b) + (a − c) = 0, so a = b = c. Conversely, it is readily checked that ifa = b = c, then equality holds.

Editor’s Comments. Using convexity and Jensen’s Inequality, Stadler proved that

in general,∑ Ä a

2b+c

äk≥ 31−k for all k ≥ 0.

4026. Proposed by Roy Barbara.

Prove or disprove the following property: if r is any non-zero rational number,then the real number x = (1 + r)1/3 + (1− r)1/3 is irrational.

We received two correct solutions. We present the solution by Joseph DiMuro.

Assume both r and x are rational numbers with r 6= 0. Setting y1 = (1 + r)1/3

and y2 = (1− r)1/3 we can show that

x3 − 2

3x= y1y2.

That means that y1 and y2 are the two solutions for y of y2− xy+ x3−23x = 0. But

using the quadratic formula, we also obtain

y =x±»x2 − 4x3−8

3x

2=x

2±…

8− x312x

.

This shows that if x is rational then y1 and y2 are contained in quadratic extensionsof Q. On the other hand, if r is rational then y1 = (1 + r)1/3 and y2 = (1− r)1/3are contained in cubic extensions of Q as well. Both of these can only be true ify1 and y2 are rational numbers themselves.

Let r = ab , where a, b are relatively prime non-zero integers. Then y1 = ( b+ab )1/3

and y2 = ( b−ab )1/3. The fractions b+ab and b−a

b are in lowest terms, so for themto be perfect cubes, their numerators and denominators must be perfect cubes.Then we have an arithmetic progression b− a, b, b+ a of cubes, which is known tobe impossible (e.g. see P. Denes, Uber die Diophantische Gleichung xl + yl = czl,Acta. Math. 88 (1952) 241-251).

Editor’s Comments. The statement that there is no arithmetic progression of threecubes can be proven with elementary number theory and is an interesting exercise.


Let a, b and c be positive real numbers such that a+ b+ c = 3. Prove that

ab

a+ ab+ b+

bc

b+ bc+ c+

ac

a+ ac+ c≤ 1.

We received 24 submissions of which 22 were correct and complete. We present 5solutions, each of them insightful in a different way.


SOLUTIONS /133

Solution 1, by Ali Adnan.

Observe that ∑cyc

ab

a+ ab+ b≤ 1 ⇐⇒

∑cyc

9a+bab + 1

≤ 9. (1)

Now, from Cauchy-Schwarz Inequality,

9a+bab + 1

=9

1a + 1

b + 1≤ a+ b+ 1,

and adding up analogous such inequalities cyclically, (1) follows.


We note that the inequality is equivalent to∑cyc

a+ b

a+ b+ ab≥ 2 ⇐⇒

∑cyc

1

2 + 2aba+b

≥ 1,

which follows easily from the AM-HM and Cauchy-Schwarz Inequalities:∑cyc

1

2 + 2aba+b

≥∑cyc

1

2 + a+b2

≥(1 + 1 + 1)2

6 + a+ b+ c= 1,

thus completing the proof.

Solution 3, by Henry Ricardo.

We have∑cyclic

ab

a+ ab+ b=

∑cyclic

11b + 1 + 1

a

=1

3

∑cyclic

31b + 1 + 1

a

≤ 1

3

∑cyclic

a+ b+ 1

3=

1

3

Å2(a+ b+ c) + 3

3

ã= 1,

where we have used the harmonic mean-arithmetic mean inequality.

Equality holds if and only if a = b = c = 1.

Solution 4, by Salem Malikic.

Using the inequality between arithmetic and geometric mean for positive reals xand y we have

x+ xy + y ≥ 3 3√x2y2

with equality if and only if x = xy = y, that gives x = y = 1 and implying that

xy

x+ xy + y≤

3√xy

3.


134/ SOLUTIONS

Using this inequality we have

ab

a+ ab+ b+

bc

b+ bc+ c+

ca

c+ ca+ a≤

3√ab+ 3

√bc+ 3

√ca

3.

Now, using Power-mean inequality, we have

3√ab+ 3

√bc+ 3

√ca

3≤ 3

…ab+ bc+ ca

3≤

3

√(a+b+c)2

3

3= 1.

where in the last step we used the well known inequality

3(ab+ bc+ ca) ≤ (a+ b+ c)2.

This completes our proof.

In order to achieve equality we must have a = b = c = 1. It is easy to verify thatthis is indeed an equality case.

Solution 5, by Leonard Giugiuc.

The inequality is equivalent to

11a + 1

b + 1+

11b + 1

c + 1+

11c + 1

a + 1≤ 1.

By AM-HM Inequality,

1

a+

1

b≥ 4

a+ b,

1

b+

1

c≥ 4

b+ c,

1

c+

1

a≥ 4

c+ a.

From here,

11a + 1

b + 1+

11b + 1

c + 1+

11c + 1

a + 1≤ 1

4a+b + 1

+1

4b+c + 1

+1

4c+a + 1

.

But

14a+b + 1

+1

4b+c + 1

+1

4c+a + 1

=a+ b

a+ b+ 4+

b+ c

b+ c+ 4+

c+ a

c+ a+ 4.

Since the function f(x) =x

x+ 4is concave if x > 0, then by Jensen’s Inequality

we get

f(a+ b) + f(b+ c) + f(c+ a) ≤ 3f

Å2(a+ b+ c)

3

ã= 3f(2) = 1.

So,a+ b

a+ b+ 4+

b+ c

b+ c+ 4+

c+ a

c+ a+ 4≤ 1.


SOLUTIONS /135


In 3-dimensional Euclidean space, a line ` meets orthogonally two distinct parallelplanes P and P ′ at H and H ′. Let r and r′ be positive real numbers with r ≤ r′;let C be the circle in P with center H, radius r, and let C′ in P ′ be similarlydefined. For a fixed point M ′ on C′, find the maximum distance between the lines` and MM ′ as M moves about the circle C (where the distance between two linesis the minimum distance from a point of one line to a point of the other).

We received four correct solutions and will feature two of them that are quite similarexcept that the first makes use of coordinates.

Solution 1, by Oliver Geupel.

We prove that the required maximum distance is r. We use Cartesian coordinatessuch that H ′ = (0, 0, 0), M ′ = (r′, 0, 0), and H = (0, 0, h) where h ∈ R. Forevery point M on C, the distance between ` and MM ′ is not greater than |MH| =r (because that distance is, by definition, the length of the shortest among allsegments joining a point of ` to a point of MM ′, which is therefore at most|MH|). Moreover, the distance between two non-intersecting lines is measuredalong a line that is perpendicular to both. Put

M =

Çr

r′· r,…

1− r2

r′2· r, h

å,

which is on C. We have−−→HM ·

−−→HH ′ = 0 and

−−−→M ′M ·

−−→HM =

Çr2

r′− r′,

…1− r2

r′2· r, h

å·Çr2

r′,

…1− r2

r′2· r, 0

å= 0,

so that HM is perpendicular to both ` and MM ′. Therefore the distance betweenthe lines ` and MM ′ is |HM | = r, which completes the proof.

Solution 2, by Edmund Swylan.

For every point M on C, let Q be the plane orthogonal to ` that contains a pointP of MM ′ nearest to `. Let O,D,D′, N,N ′ be the orthogonal projections of`, C, C′,M,M ′, respectively, onto Q. The distance between the lines ` and MM ′

projects to |PO|. Our problem is thereby reduced to a 2-dimensional problem:

Given circles D and D′ in the same plane with common centre O andradii r and r′, a fixed point N ′ on D′, and a point N moving about D,what is the maximum distance from O to NN ′?

The answer is r.

For r < r′ the maximum is achieved if and only if NN ′ is tangent to D. For r = r′

it is achieved if and only if N = N ′ and the line NN ′ degenerates into a pointwhich occurs when MM ′ is parallel to `.


136/ SOLUTIONS

4029. Proposed by Paul Bracken.

Suppose a > 0. Find the solutions of the following equation in the interval (0,∞):

1

x+ 1+∞∑n=1

n!

(x+ 1)(x+ 2) · · · (x+ n+ 1)= x− a.

We received four correct solutions and will feature two different ones.

Solution 1. We present a composite of the very similar solutions by Arkady Altand the proposer, Paul Bracken. Another similar solution was received from OliverGeupel.

It is clear that

1

x− 1

x+ 1=

1

x(x+ 1),

1

x− 1

x+ 1− 1

(x+ 1)(x+ 2)=

2

x(x+ 1)(x+ 2),

and

n!

x(x+ 1)(x+ 2) · · · (x+ n)− n!

(x+ 1)(x+ 2) · · · (x+ n+ 1)

=(n+ 1)!

x(x+ 1)(x+ 2) · · · (x+ n+ 1).

It therefore follows by induction that

1

x− 1

x+ 1−n−1∑k=1

k!

(x+ 1)(x+ 2) . . . (x+ k + 1)=

n!

x(x+ 1)(x+ 2) . . . (x+ n).

However, for x > 0,

limn→∞

n!

x(x+ 1) . . . (x+ n)= 0,

sincen!

x(x+ 1)(x+ 2) · · · (x+ n)=

1

x (x+ 1)(x2 + 1

)· · ·(xn + 1

)and

(x+ 1)(x

2+ 1)· · ·(xn

+ 1)> 1 + x

Å1 +

1

2+ · · ·+ 1

n

ã.

Hence the left-hand side of the original equation is given by

1

x+ 1+∞∑n=1

n!

(x+ 1)(x+ 2) · · · (x+ n+ 1)=

1

x.

Therefore the original equation is equivalent to x2 − ax − 1 = 0. This quadraticequation has the following unique solution in (0,∞):

xr =1

2(a+

√a2 + 4).


SOLUTIONS /137

Solution 2, by Albert Stadler.

We note that

∞∑n=1

n!

(x+ 1)(x+ 2) · · · (x+ n+ 1)

=∞∑n=1

Γ(x+ 1)Γ(n+ 1)

Γ(x+ 1)(x+ 2) · · · (x+ n+ 1)

=∞∑n=1

Γ(x+ 1)Γ(n+ 1)

Γ(x+ n+ 1)

=∞∑n=1

β(x+ 1, n+ 1)

=∞∑n=1

∫ 1

0

tx(1− t)n dt

=

∫ 1

0

tx−1(1− t) dt

=1

x− 1

x+ 1, x > 0.

The original equation is therefore equivalent to 1x = x−a or x2−ax−1 = 0. This

quadratic equation has exactly one positive root, which is xr = 12 (a+

√a2 + 4).

4030. Proposed by Paolo Perfetti.

a) Prove that 4cos t + 4sin t ≥ 5 for t ∈ [0, π4 ].

b) Prove that 6cos t + 6sin t ≥ 7 for t ∈ [0, π4 ].

There were six submitted solutions for this problem, four of which were correct.We present the solution by Michel Bataille.

Lemma. Let u(t) = sin t cos t(cos t + sin t). Then, u is an increasing function on

[0, π4 ] with u(0) = 0 and u(π4 ) =√22 .

Proof. u(0) = 0, u(π4 ) =√22 are immediate. A simple calculation gives the deriva-

tive of u:u′(t) = (cos t− sin t)((cos t+ sin t)2 + sin t cos t).

For t ∈ (0, π4 ), cos t > sin t, hence u′(t) > 0 and so u is increasing on [0, π4 ]. 2

a) Let f(t) = 4cos t + 4sin t. We show that f is increasing on [0, π4 ] (the requiredresult then follows since f(0) = 5). To this end, we prove that f ′(t) > 0 for allt ∈ (0, π4 ). We calculate

f ′(t) = (ln 4)4cos t cos t

Å4sin t−cos t − sin t

cos t

ãCopyright c© Canadian Mathematical Society, 2016

138/ SOLUTIONS

so that it is sufficient to prove that φ(t) > 0 for t ∈ (0, π4 ) where

φ(t) = (sin t− cos t)(ln 4)− ln(sin t) + ln(cos t).

Now, we easily obtain φ′(t) = (ln 4)u(t)−1sin t cos t with, from the lemma,

(ln 4)u(t)− 1 <

√2 ln 4

2− 1 < 0.

Therefore, φ′(t) < 0 for t ∈ (0, π4 ) and φ(t) > φ(π4

)= 0, as desired.

b) Similarly, we introduce g(t) = 6cos t + 6sin t whose derivative has the same signas ψ(t) = (sin t− cos t)(ln 6)− ln(sin t) + ln(cos t). Here,

ψ′(t) =ln 6

sin t cos t·Åu(t)− 1

ln 6

ã,

and since 0 < 1ln 6 <

√22 , u(t)− 1

ln 6 (and so ψ′(t)) vanishes at a unique t0 in (0, π4 ).From the lemma, we deduce that ψ′(t) < 0 if 0 < t < t0 and ψ′(t) > 0 if t0 < t < π

4 .Thus, ψ is decreasing on (0, t0] and increasing on [t0,

π4 ). Since ψ(π4 ) = 0, we must

have ψ(t0) < 0, and since limt→0+

ψ(t) =∞, we deduce that for some α ∈ (0, t0), we

have ψ(t) > 0 if t ∈ (0, α), ψ(α) = 0 and ψ(t) < 0 if t ∈ (α, π4 ). Thus, g′(t) > 0 ift ∈ (0, α) and g′(t) < 0 if t ∈ (α, π4 ) and so g(t) ≥ (min(g(0), g(π/4)) = 7 for allt ∈ [0, π4 ].

Editor’s Comments. It turns out that AM-GM is too weak to prove this inequal-ity when used right at the beginning; the resulting right-hand-side is too small.However, one may use AM-GM in a step of the proof, as A. Stadler did, and havethings work out well; the Stadler solution is an impressive use of Taylor series andclever bounds. As well, the ‘general’ inequality, acos(t) + asin(t) ≥ a + 1, is nottrue over the required interval for every a > 1; plotting it for a = 10, for example,shows this.


176/ SOLUTIONS



4031. Proposed by D. M. Batinetu-Giurgiu and Neculai Stanciu.

Prove that

2F 41 + F 4

2 + F 43

F 21 + F 2

3

+2F 4

2 + F 43 + F 4

4

F 22 + F 2

4

+ · · ·+ 2F 4n + F 4

1 + F 42

F 2n + F 2

2

> 2FnFn+1,

where Fn represents the nth Fibonacci number (F0 = 0, F1 = 1 and Fn+2 =Fn + Fn+1 for all n ≥ 1).

We received five correct solutions. We present two solutions.

Editor’s comments. When n = 1, the interpretation of the left side is not clear,while when n = 2, we obtain equality. Therefore, we suppose that n ≥ 3.

Solution 1, by Adnan Ali and the proposers (independently).

Observe that, for positive x, y, z,

2x2 + y2 + z2

x+ z≥ x+ y

with equality if and only if x = y = z, since this inequality is equivalent to(x− y)2 + (y − z)2 + (z − x)2 ≥ 0. It follows that the left side of the inequality isgreater than

(F 21 + F 2

2 ) + (F 22 + F 2

3 ) + · · ·+ (F 2n−1 + F 2

n) + (F 2n + F 2

1 ) = 2n∑k=1

F 2k .

Since∑nk=1 F

2k = FnFn+1 (easily obtained by induction for n ≥ 1), the result

follows.


For positive x, y, z,x2

y + z≥ 4x− y − z

4

with equality iff 2x = y + z. This implies that

2F 4i + F 4

j + F 4k

F 2i + F 2

k

>2(4F 2

i − F 2i − F 2

k ) + (4F 2j − F 2

i − F 2k ) + (4F 2

k − F 2i − F 2

k )

4

= F 2i + F 2

j ,

Crux Mathematicorum, Vol. 42(4), April 2016

SOLUTIONS /177

for distinct i, j, k (since not all of Fi, Fj , Fk are equal to Fi + Fk). Thus the leftside is greater than 2

∑nk=1 F

2k = 2FnFn+1.

4032. Proposed by Dan Stefan Marinescu and Leonard Giugiuc.

Prove that in any triangle ABC with sides a, b and c, inradius r and exradiira, rb, rc, we have:

√ab+

√bc+

√ca ≥ 2

»3r(ra + rb + rc).

We received 13 correct solutions. We present two solutions.

Solution 1, by Titu Zvonaru.

Using Ravi’s substitutions (a = y + z, b = z + x, c = x + y, with x, y, z > 0), wehave

[ABC] =»xyz(x+ y + z), r =

…xyz

x+ y + z, ra =

√xyz(x+ y + z)

x,

so that

r(ra + rb + rc) = xyz

Å1

x+

1

y+

1

z

ã= xy + yz + zx.

We have to prove that»(x+ y)(y + z) +

»(y + z)(z + x) +

»(z + x)(x+ y) ≥ 2

»3(xy + yz + zx).

Using Minkowski’s inequality and the inequality (x + y + z)2 ≥ 3(xy + yz + zx),we obtain»

(x+ y)(y + z) +»

(y + z)(z + x) +»

(z + x)(x+ y)

=»x2 + (xy + yz + zx) +

»y2 + (xy + yz + zx) +

»z2 + (xy + yz + zx)

≥»

(x+ y + z)2 + (3√xy + yz + zx)2

≥»

3(xy + yz + zx) + 9(xy + yz + zx)

= 2»

3(xy + yz + zx).

Equality holds if and only if x = y = z; that is, if and only if triangle ABC isequilateral.

Solution 2, composite of similar solutions by Sefket Arslanagic and Kee-Wai Lau.

By the known equalityra + rb + rc = 4R+ r,

the given inequality is equivalent to

√ab+

√bc+

√ca ≥ 2

»3r(4R+ r).


178/ SOLUTIONS

By the AM-GM inequality,√ab+

√bc+

√ca ≥ 3 · 3

√abc.

It therefore suffices to show that

3 · 3√abc ≥ 2 ·

»3r(4R+ r),

which is successively equivalent to

3 · 3√

4Rrs ≥ 2 ·»

3r(4R+ r)

36 · 16R2r2s2 ≥ 26 · 27r3(4R+ r)3

27R2s2 ≥ 4r(4R+ r)3.

By the inequalities s2 ≥ 16Rr − 5r2 due to J.C. Gerretsen and R ≥ 2r due to L.Euler, we have

27R2s2 − 4r(4R+ r)3 ≥ 27(16Rr − 5r2)R2 − 4r(4R+ r)3

= r(R− 2r)(176R2 + 25Rr + 2r2)

≥ 0.

This proves the inequality of the problem. Equality holds for the equilateral tri-angle.

4033. Proposed by Salem Malikic.

Let α1, . . . , αn, β1, . . . , βn be positive real numbers and x1, . . . , xn be real numberssuch that x1 + · · · + xn = 1 and αixi + βi ≥ 0 for all i = 1, . . . , n. Find themaximum value of√

α1x1 + β1 +√α2x2 + β2 + · · ·+

√αnxn + βn.

We received eight correct solutions and one incorrect solution. We present a com-posite of three nearly identical solutions given independently by Adnan Ali, JoeSchlosberg, and Titu Zvonaru.

Let M denote the required maximum value. Applying the Cauchy-Schwarz In-equality, we have√

α1x1 + β1 +√α2x2 + β2 + · · ·+

√αnxn + βn

=√α1

x1 +

β1α1

+√α2

x2 +

β2α2

+ · · ·+√αn

xn +

βnαn

≥

(α1 + α2 + · · ·+ αn)(x1 + x2 + · · ·+ xn +β1α1

+β2α2

+ · · ·+ βnαn

)

=

(α1 + α2 + · · ·+ αn)(1 +

β1α1

+β2α2

+ · · ·+ βnαn

).


SOLUTIONS /179

The equality holds if and only if for some k we have

x1 + β1

α1

α1=x2 + β2

α2

α2= · · · =

xn + βn

αn

αn= k.

Since

k =

∑xi +

∑βi

αi∑αi

=1 +

∑βi

αi∑αi

,

we have

xi = αi ·1 +

∑βi

αi∑αi

− βiαi, for i = 1, 2, . . . , n. (1)

Therefore,

M =

(α1 + α2 + · · ·+ αn)(1 +

β1α1

+β2α2

+ · · ·+ βnαn

)

attained when xi are as in (1).


Evaluate∞∑n=0

(−1)n

16n

2n∑k=0

(−1)k

2n+ 1− k

Ç2k

k

åÇ4n− 2k

2n− k

å.

We received three correct solutions. We present the solution by Angel Plaza.

Let us denote

an =n∑k=0

(−1)k

n+ 1− k

Ç2k

k

åÇ2n− 2k

n− k

å,

so the proposed expression reads as

∞∑n=0

a2n

Å−1

16

ãn.

We will use the snake oil method to find the generating function of the sequencewith general term an. If F (x) is its generating function then


180/ SOLUTIONS

F (x) =∑n≥0

anxn

=∑n≥0

(n∑k=0

(−1)k

n+ 1− k

Ç2k

k

åÇ2n− 2k

n− k

å)xn

=∑k≥0

(−1)kÇ

2k

k

å∑n≥k

1

n+ 1− k

Ç2n− 2k

n− k

åxn

=∑k≥0

(−1)kÇ

2k

k

åxk∑n≥0

1

n+ 1

Ç2n

n

å=∑k≥0

Ç2k

k

å(−x)k

2

1 +√

1− 4x

=1√

1 + 4x· 2

1 +√

1− 4x,

where we used the generating functions of the Catalan number and of the centralbinomial coefficients (both with radius of convergence |x| < 1

4 ) in the last twosteps. Now ∑

n≥0a2nx

n =F (√x) + F (−

√x)

2,

so the proposed sum is equal to

1√1 + i

· 1

1 +√

1− i+

1√1− i

· 1

1 +√

1 + i,

which can be calculated to√

2−√√

2− 1.

4035. Proposed by Daniel Sitaru and Leonard Giugiuc.

Let a and b be two real numbers such that ab = 225. Find all real solutions (inreal 2× 2 matrices) to the matrix equation

X3 − 5X2 + 6X =

Å15 ab 15

ã.

We received four submissions for this question, of which three were correct andcomplete. We present the solution by Michel Bataille.

We will show that the solutions are the three matrices

X1 =

Å5/2 a/6b/6 5/2

ã, X2 =

Å4 a/15

b/15 4

ã, X3 =

Å7/2 a/10b/10 7/2

ã.

Simple calculations show that these three matrices satisfy the given equation.


SOLUTIONS /181

Let A =

Å15 ab 15

ãand let I2 be the 2 × 2 unit matrix. Since det(xI2 − A) =

x2 − 30x, the eigenvalues of A are 0 and 30.

Let X be a solution of the given equation and let λ be an eigenvalue of X. Thenλ3 − 5λ2 + 6λ is an eigenvalue of A, hence λ3 − 5λ2 + 6λ ∈ {0, 30}. Thus,

λ(λ− 2)(λ− 3) = 0 or (λ− 5)(λ2 + 6) = 0

and the possible eigenvalues of X are 0, 2, 3, 5, i√

6,−i√

6.

Noting that the characteristic polynomial χ(x) of the real matrix X is in R[x], ifi√

6 (resp. −i√

6) is an eigenvalue of X, so is its complex conjugate and then X

is similar to

Åi√

6 0

0 −i√

6

ã. But then A = X3 − 5X2 + 6X is similar toÅ

−6i√

6 0

0 6i√

6

ã− 5

Å−6 00 −6

ã+ 6

Åi√

6 0

0 −i√

6

ã=

Å30 00 30

ã,

a contradiction since 0 is an eigenvalue of A.

As a result, if λ1, λ2 are the eigenvalues of X (possibly λ1 = λ2), we have λ1, λ2 ∈{0, 2, 3, 5}. Since the eigenvalues of X are real, X is similar to an upper triangularreal matrix, say,

X = P

Åλ1 α0 λ2

ãP−1

for some invertible real matrix P and some real number α. Then

A = X3 − 5X2 + 6X = P

Åλ31 − 5λ21 + 6λ1 β

0 λ32 − 5λ22 + 6λ2

ãP−1

where β ∈ R. Taking traces, it follows that

30 = (λ31 − 5λ21 + 6λ1) + (λ32 − 5λ22 + 6λ2) (1).

After trying all possibilities for λ1 and λ2, we realize that λ1 6= λ2 and {λ1, λ2} iseither {0, 5} or {2, 5} or {3, 5}. We consider the three cases in turn:

• if {λ1, λ2} = {0, 5}, then χ(x) = x2 − 5x. Note that

x3 − 5x2 + 6x = x(x2 − 5x) + 6x = xχ(x) + 6x.

By the Cayley-Hamilton theorem, χ(X) = 0; replacing x by X in the above

we get A = Xχ(X) + 6X = 6X, hence X = 16A =

Å5/2 a/6b/6 5/2

ã.

• if {λ1, λ2} = {2, 5}, then χ(x) = (x− 2)(x− 5) and

x3 − 5x2 + 6x = (x+ 2)χ(x) + 10x− 20.

Therefore A = 10X − 20I2 and

X =1

10(A+ 20I2) =

Å7/2 a/10b/10 7/2

ã.


182/ SOLUTIONS

• if {λ1, λ2} = {3, 5}, then χ(x) = (x− 3)(x− 5) and

x3 − 5x2 + 6x = (x+ 3)χ(x) + 15x− 45.

Hence A = 15X − 45, which gives us

X =1

15(A+ 45I2) =

Å4 a/15

b/15 4

ã.

The proof is complete.


Let a, b and c be non-negative real numbers. Prove that for any real k ≥ 1124 we

have:

k(ab+ bc+ ca)(a+ b+ c)− (a2c+ b2a+ c2b) ≤ (3k − 1)(a+ b+ c)3

9.

We received five submissions all of which are correct. We present the solution bythe proposer, slightly modified by the editor.

Due to cyclic symmetry of the functions involved, we may assume that c =min {a, b, c} .

Let x = a − c and y = b − c . Then x, y, c ≥ 0, a = x + c, b = y + c, anda+ b+ c = x+ y + 3c.

The given inequality is equivalent to

(3k − 1) (a+ b+ c)3 − 9k (ab+ bc+ ca) (a+ b+ c) + 9

(a2c+ b2a+ c2b

)≥ 0

or

(3k − 1)(x+ y + 3c)3 − 9k((x+ c)(y + c) + c(x+ y + 2c))(x+ y + 3c)

+ 9((x+ c)2c+ (y + c)2(x+ c) + c2(y + c)) ≥ 0. (1)

Let F (x, y, c) denote the expression on the left hand side of (1), and set p = x+ yand q = xy.Since 9k((x+ c)(y+ c) + c(x+ y+ 2c))(x+ y+ 3c) = 9k(3c2 + 2pc+ q)(p+ 3c) and

9((x+ c)2c+ (y + c)2(x+ c) + c2(y + c))

= 9(cx2 + cy2 + 2cxy + 3c2(x+ y) + xy2 + 3c3)

= 9(cp2 + 3c2p+ 3c3 + xy2)

= 9cp2 + 27c2p+ 27c3 + 9xy2

= (p+ 3c)3 − p3 + 9xy2,


SOLUTIONS /183

we have

F (x, y, c) = (3k − 1)(p+ 3c)3 − 9k(3c2 + 2pc+ q)(p+ 3c) + (p+ 3c)3 − p3 + 9xy2

= 3k(p+ 3c)3 − 9k(3c2 + 2pc+ q)(p+ 3c)− p3 + 9xy2

= 3k(p3 + 9cp2 + 27c2p+ 27c3)− 9k(2cp2 + 9c2p+ 9c3 + pq + 3cq)

− p3 + qxy2

= (3k − 1)p3 + 9ckp2 − 27ckq − 9kpq + 9xy2

= (3k − 1)(x+ y)3 + 9ck(x+ y)2 − 27ckxy − 9kxy(x+ y) + 9xy2

= (3k − 1)(x3 + y3 + 3xy(x+ y)) + 9ck(x2 + 2xy + y2)− 27ckxy

− 9kxy(x+ y) + 9xy2

= (3k − 1)x3 + 6xy2 − 3x2y + 9ck(x2 − xy + y2) + (3k − 1)y3. (2)

Clearly, 9ck(x2 − xy + y2) + (3k − 1)y3 ≥ 0. Furthermore,

(3k − 1)x3 + 6xy2 − 3x2y = x((3k − 1)x2 − 3xy + 6y2) ≥ 0

since the discriminant of (3k − 1)x2 − 3xy + 6y2 is

9y2 − 24(3k − 1)y2 = 3(11− 24k)y2 ≤ 0

and 3k − 1 > 0. Hence, from (2) we conclude that F (x, y, c) ≥ 0 which by (1)completes the proof.


Let P be a point of the incircle γ of a triangle ABC. The perpendiculars toBC,CA and AB through P meet γ again at U, V and W , respectively. Prove thatthe area of UVW is independent of the chosen point P on γ.

We received six correct and complete solutions. We present the solution by OliverGeupel. Ricard Peiro and Prithwijt De submitted similar solutions.

We prove that triangle UVW is similar to triangle ABC. As a consequence, sinceγ is the circumcircle of UVW , the area of triangle UVW is

[UVW ] =r2

R2[ABC],

where r and R denote the inradius and the circumradius, respectively, of 4ABC.

Let A, B, C, U , V , and W denote measures of the the interior angles of thetriangles ABC and UVW . Since PV ⊥ AC and PW ⊥ AB, the size of ∠V PW is180◦ − A. Also, since the points P , U , V , and W are concyclic, ∠V UW is equalto either ∠V PW or 180◦ − ∠V PW . Hence, U ∈ {A, 180◦ − A}. Analogously,V ∈ {B, 180◦ − B} and W ∈ {C, 180◦ − C}.

We show that (U , V , W ) = (A, B, C).


184/ SOLUTIONS

Assume the contrary. Then, there is no loss of generality in assuming that U 6= A.Thus, U = 180◦ − A.

If (V , W ) = (B, C), then we obtain 180◦ = U + V + W = 180◦ − A + B + C,so that A = B + C = 90◦ = 180◦ − A = U , contradicting our assumption thatU 6= A. Hence, (V , W ) 6= (B, C). There is no loss of generality in assuming thatV = 180◦−B. But then, U+ V = (180◦−A)+(180◦−B) > 180◦, a contradiction.

Editor’s Comments. We were pleased to find that among the six solutions sub-mitted, four different formulas for the area of a triangle were used.


Let x, y, z be positive real numbers such that x+ y+ z = xyz. Find the minimumvalue of the expression…

1

3x4 + 1 +

…1

3y4 + 1 +

…1

3z4 + 1.

There were 21 correct solutions, with four from one submitter and three fromanother. An additional solution was incorrect.

Solution 1, by Arkady Alt, Sefket Arslanagic, and Daniel Dan (independently).

Since xyz = x+y+z ≥ 3 3√xyz, if follows that x+y+z = xyz ≥ 3

√3. Applying the

inequality of the root mean square and arithmetic mean, we have, for t = x, y, z,…1

3t4 + 1 =

√Åt2

3

ã2+

Åt2

3

ã2+

Åt2

3

ã2+ 1

≥ (t2/3) + (t2/3) + (t2/3) + 1

2=t2 + 1

2

with equality iff t =√

3. (Alternatively, the inequality»

13 t

4 + 1 ≥ t2+12 is equiv-

alent to (t2 − 3)2 ≥ 0.) Therefore, the left side of the inequality is not less than

1

2(3 + x2 + y2 + z2) ≥ 1

2

Å3 +

(x+ y + z)2

3

ã≥ 1

2(3 + (27/3)) = 6.

Since equality occurs when x = y = z =√

3, the desired minimum is 6.


SOLUTIONS /185

Solution 2, by Sefket Arslanagic and Salem Malikic (independently).

As before, x+ y + z ≥ 3√

3. We begin by establishing that…1

3u4 + 1 ≥ u

√3− 1,

with equality iff u =√

3. The result is clear when u < 1/√

3. When u ≥ 1/√

3,the inequality is equivalent to

1

3u4 + 2

√3u ≥ 3u2.

By the arithmetic-geometric means inequality, we obtain that

1

3u4 + 2

√3u =

1

3u4 + u

√3 + u

√3 ≥ 3

3

…1

3u4 · u

√3 · u√

3 = 3u2

as desired.

The left side of the inequality of the problem is not less than√

3(x+ y + z)− 3 = 9− 3 = 6,

with equality iff x = y = z =√

3. The desired minimum is 6.


From the triangle inequality in Euclidean space R2,

‖a + b + c‖2 ≤ ‖a‖2 + ‖b‖2 + ‖c‖2,

applied to a = (x2/√

3, 1), b = (y2/√

3, 1), c = (z2/√

3, 1), we have that…x4

3+ 1 +

…y4

3+ 1 +

…z4

3+ 1 ≥

√Åx2√

3+y2√

3+

z2√3

ã2+ (1 + 1 + 1)2

=

(x2 + y2 + z2)2

3+ 9,


3. Using x2 + y2 + z2 ≥ xy + yz + zx, thearithmetic-harmonic means inequality and x+y+z = xyz in turn, we obtain that

x2 + y2 + z2 ≥ xy + yz + zx ≥ 9xyz

x+ y + z= 9.

Thus the left side of the inequality is not less than…81

3+ 9 = 6,


3. The desired minimum is 6.


186/ SOLUTIONS


By the Cauchy-Schwarz inequality,…1

3t4 + 1 ·

√3 + 1 ≥ t2 + 1,

and by the arithmetic-geometric means inequality,

(x2 + y2 + z2)(x+ y + z) ≥ 3(xyz)2/3 · 3(xyz)1/3 = 9xyz.

Therefore …1

3x4 + 1 +

…1

3y4 + 1 +

…1

3z4 + 1 ≥ 1

2(x2 + y2 + z2) +

3

2

≥ 1

2

Å9xyz

x+ y + z

ã+

3

2= 6,

with equality if and only if x = y = z =√

3.

Editor’s comments. Seven solvers applied Jensen’s Inequality to obtain the result,the function

√(x4/3) + 1 being convex. Some solvers noted that, under the stated

constraint, we can write (x, y, z) = (tanα, tanβ, tan γ) with α + β + γ = π andeach angle less than π/2, or (x, y, z) = (cotλ+cotµ+cot ν) with λ+µ+ν = π/2,and then obtain a trigonometric inequality.


In a triangle ABC, let ∠CAB = 48◦ and ∠CBA = 12◦. Suppose D is a point onAB such that CD = 1 and AB =

√3. Find ∠DCB.

We received 17 correct solutions and will feature the solution submitted by theSkidmore College Problem Group.

Let ∠DCB = γ; we shall show that γ = 6◦.

By the Law of Sines applied to triangle ABC, sin 12◦

AC = sin 120◦√3

= 12 , so we have

AC = 2 sin 12◦.

For triangle ACD, sin(12◦+γ)AC = sin 48◦

1 , whence

sin(12◦ + γ) = 2 sin 48◦ sin 12◦ = cos 36◦ − cos 60◦ = cos 36◦ − 1

2.

Let φ = 1+√5

2 (= φ2 − 1) be the golden section (that is, the ratio of a diagonal ofthe regular pentagon to a side). We know that

sin 18◦ =1

2φ=φ

2− 1

2and cos 36◦ =

φ

2.

[For example, if the regular pentagon PQRST has unit sides, then the isoscelestriangle PRS has apex angle 36◦ and sides φ and 1 (which provides the value for


SOLUTIONS /187

sin 18◦), while the isosceles triangle PQT with base angle 36◦ also has sides 1 andφ (which gives cos 36◦).] Thus cos 36◦ = sin 18◦ + 1

2 , and

sin(12◦ + γ) = sin 18◦ = sin 162◦.

But γ < 120◦, which implies that γ = 6◦, as claimed.

Editor’s comments. In a remark added to his solution, Dag Jonsson observed thatthere would be a second solution should we allow D to be a point of the line AB(instead of restricting it to the segment AB). Vertex A would then lie between Dand B; the argument of the featured solution remains valid so that ∠ADC = 18◦,whence ∠DCA = 30◦ and, finally, ∠DCB = 150◦.

Returning to the case where D lies between A and B, we note that the degeneratequadrangle ADBC of our problem is an example of an adventitious quadrangle.Adventitious quadrangles were first defined by Colin Tripp [5], but his narrowdefinition was later extended to include any quadrangle for which the angles formedby sides and diagonals are all rational multiples of 180◦. Rigby [3] observed thatthe problem of classifying the adventitious quadrangles is equivalent to classifyingall triple intersections of diagonals in regular polygons, a problem solved manyyears earlier by Gerrit Bol [1]. A complete account, including an elementarysummary and a 15-item bibliography, was provided by Poonen and Rubinsteinin [2]. Triangle ABC of our problem can be inscribed in a regular 30-gon thathas a unit circumradius. Label its vertices from 0 to 29 and place A at 10, Bat 0, and C at 8. Note that the angle subtended at a vertex of the 30-gon byany nonadjacent edge is 6◦, which immediately implies that the angles of ∆ABCare indeed 48◦, 12◦, and 120◦. Because AB is the edge of an inscribed equilateraltriangle (with vertices numbered 0, 10, 20), we have AB =

√3 as desired. Let D′

be the point of intersection of the diameter 2, 17 and the diagonal 8, 29. If O isthe circumcentre, then triangle OD′C is isosceles (with angles 36◦, 72◦, and 72◦),whence CD′ = CO = 1. It remains to prove that D′ ∈ AB, which will immediatelyimply that D′ = D (and, consequently, that ∠DCB = ∠D′CB = 6◦). On p. 223of [4] Rigby declares (with a slightly different labeling) that the diagonals 29, 8;0, 10; and 2, 17 are indeed concurrent. While the author provides a “geometric”proof that these three lines are concurrent, it is perhaps more efficient to usetrigonometry. Applying the sine form of Ceva’s theorem to the triangle whosevertices are those numbered 0, 8, 17 with cevians joining 0 to 10, 8 to 29, and 17to 2 (as in the accompanying figure), we must show that the product

sin 42

sin 12· sin 6

sin 72· sin 36

sin 12

equals 1. This is easily accomplished using an argument similar to that of ourfeatured solution.


188/ SOLUTIONS

References

[1] G. Bol, Beantwoording van prijsvraag no. 17, Nieuw archief voor Wiskunde,18 (1936) 14-66.

[2] Bjorn Poonen and Michael Rubinstein, The Number of Intersection PointsMade by the Diagonals of a Regular Polygon. Siam Journal of Discrete Math-ematics 11:1 (February 1998) 135-156.

[3] J.F. Rigby, Adventitious Quadrangles: A Geometrical Approach. MathematicalGazette 62:421 (October 1978) 183-191.

[4] J.F. Rigby, Multiple Intersections of Diagonals of Regular Polygons and Re-lated Topics. Geometriae Dedicata 9:2 (1980) 207-238.

[5] C.E. Tripp, Adventitious Angles. Mathematical Gazette 59:408 (June 1975)98-106.


SOLUTIONS /189

4040. Proposed by Ali Behrouz.

Find all functions f : N 7→ N such that

(f(a) + b)f(a+ f(b)) = (a+ f(b))2 ∀a, b ∈ N.

There were nine submitted solutions for this problem, all of which were correct. Wepresent three solutions; the first solution is for the case that 0 is the first naturalnumber, and the latter two are for the case that 1 is the first natural number.


Suppose that 0 is the first natural number. Let a = b = 0; we have:

f(0)f(f(0)

)= f(0)2,

which implies that either f(0) = 0 or f(f(0)

)= f(0). If f

(f(0)

)= f(0), then we

replace a with 0 and b with f(0) in the given relation and obtain:

2f(0)f(0) = f(0)2,

which implies that f(0)2 = 0. Hence no matter what, f(0) = 0. Now replace bwith 0 and obtain:

f(a)2 = a2,

for all a ≥ 0, and so f(a) = a for all a ≥ 0, because f(a) ≥ 0. Thus the requiredfunction is the identity (which clearly satisfies the given relation).

Solution 2, by Adnan Ali.

Suppose that 1 is the first natural number. Denote by P (a, b) the above functionalequation for all a, b ∈ N. Now let f(1) = k, where k is a natural number. ThenP (1, 1) gives f(k + 1) = k + 1. Using this, P (1, k + 1) implies

(2k + 1)f(k + 2) = (k + 2)2.

This means that (2k + 1) divides (k + 2)2. This means that 2k + 1 divides

4(k + 2)2 − 8(2k + 1)− (2k + 1)(2k − 1) = 9.

This forces 2k + 1 to be either 3 or 9, which gives f(1) = 1 or f(1) = 4.

If f(1) = 4, then from P (1, 1) we get f(5) = 5. Next P (5, 1) gives

(f(5) + 1)f(5 + f(1)) = (5 + f(1))2 = 81,

implying that 6f(9) = 81, which is clearly impossible as 6 - 81 and f(9) ∈ N.

Thus we must have f(1) = 1. Now we prove by induction that f(1) = 1 impliesthat f(n) = n for all n ∈ N. The case n = 1 is already true. Now assume thatf(a) = a for some a ≥ 1. Then from P (a, 1):

(f(a) + 1)f(a+ f(1)) = (a+ f(1))2,


190/ SOLUTIONS

which implies that f(a+ 1) = a+ 1, and consequently by induction f(n) = n forall n ∈ N. So, in summary the only function satisfying the equation is the functionf(n) = n for all n ∈ N.

Solution 3, by Joseph Ling.

It is easy to see that f (n) = n for all n satisfies the required relation for all a, b ∈ N

We show that there are no other solutions.

First, we note that f is one-to-one. For if b1 and b2 are such that f (b1) = f (b2) ,then for all a, we have

f (a) + b1 =(a+ f (b1))

2

f (a+ f (b1))=

(a+ f (b2))2

f (a+ f (b2))= f (a) + b2,

implying that b1 = b2.

Next, we note that f is (strictly) increasing. For if there exist b1 < b2 withf (b1) > f (b2) , then for every a ∈ N, we consider a′ = a+f (b1)−f (b2) . We havea′ ∈ N, with a′ + f (b2) = a+ f (b1) , and consequently,

f (a′) + b2 =(a′ + f (b2))

2

f (a′ + f (b2))=

(a+ f (b1))2

f (a+ f (b1))= f (a) + b1 < f (a) + b2

implying that f (a′) < f (a) . But then this means that the range of f has nosmallest element, contradicting the well-ordering principle.

Now, since f is strictly increasing, a simple induction shows that f (n) ≥ n forall n ∈ N. It follows that f (a+ f (b)) ≥ a + f (b) for all a, b ∈ N. Consequently,the given relation implies that f (a) + b ≤ a + f (b) for all a, b ∈ N. But then bysymmetry, we conclude that in fact, f (a)+b = a+f (b) for all a, b ∈ N. Therefore,

f (n) = n+ k

for all n ∈ N, where k = f (1)− 1. Applying this to the relation, we get

(a+ b+ k) (a+ b+ 2k) = (a+ b+ k)2

for all a, b ∈ N. But then this implies that k = 0, and so, f (n) = n for all n ∈ N.

Editor’s Comments. Most of the solvers, and the proposer, assumed simply that0 was not a natural number; only Giugiuc solved the problem with both interpre-tations. Assuming that 0 is natural means there is more known information, andthus it is reasonable that the solution is easier than in the other case. Regardingthe other case, the solution by Ali uses divisibility, a number-theoretic property,whereas the solution by Ling uses the function-theoretic properties of solutions tothis relation. Other solutions use the relation differently (some use it to find addi-tivity of the function, others show that there are infinitely many primes satisfyingf(n) = n, etc.).


SOLUTIONS /225




Let a, b and c be the side lengths of a triangle ABC. Let AA′, BB′ and CC ′ bethe heights of the triangle and let ap = B′C ′, bp = C ′A′ and cp = A′B′ be thesides of the orthic triangle. Prove that:

a) a2 (bp + cp) + b2 (cp + ap) + c2 (ap + bp) = 3abc;

b) ap + bp + cp ≤ s, where s is the semiperimeter of ABC.

We received 15 correct solutions and present the solution by Michel Bataille.

We show (a) and (b) in the case when ∆ABC has no obtuse angle and provide acounter-example in the opposite case.

First, suppose that angles A, B, C are acute. Since ∆AB′B is right-angled with∠AB′B = 90◦, we have AB′ = c · cosA. Similarly, AC ′ = b · cosA, and it followsthat

B′C ′2 = c2 cos2A+ b2 cos2A− 2bc cos3A

= (c2 + b2 − 2bc cosA) cos2A = a2 cos2A

and so ap = B′C ′ = a cosA. In a similar way, we obtain bp = A′C ′ = b cosB andcp = A′B′ = c cosC.

Now we calculate X = a2(bp + cp) + b2(cp + ap) + c2(ap + bp) as follows:

X = a2b cosB + a2c cosC + b2c cosC + b2a cosA+ bc2 cosB + c2a cosA

= ab(a cosB + b cosA) + bc(b cosC + c cosB) + ca(c cosA+ a cosC)

= abc+ bca+ cab = 3abc,

as desired. Denoting by r and R the inradius and the circumradius of ∆ABC andusing the Law of Sines, we get

ap + bp + cp = a cosA+ b cosB + c cosC

= R(sin 2A+ sin 2B + sin 2C)

= 4R sinA sinB sinC

= 4R · abc8R3

=4rRs

2R2= s · 2r

R

and the result ap + bp + cp ≤ s follows from Euler’s inequality 2r ≤ R.


226/ SOLUTIONS

If ∆ABC is right-angled, say ∠BAC = 90◦, results (a) and (b) continue to holdif we take, as is natural, ap = 0, bp = cp = h, where h = AA′. Indeed, we have3abc = 3a · ah = 3a2h and

a2(bp + cp) + b2(cp + ap) + c2(ap + bp) = a2 · 2h+ b2 · h+ c2 · h= h(b2 + c2 + 2a2) = 3a2h.

Also, the inequality ap+bp+cp ≤ s rewrites as 4h ≤ a+b+c or 4bc ≤ a2+a(b+c).

Since b+ c ≥ 2√bc and a2 = b2 + c2 ≥ 2bc, we have

a2 + a(b+ c) ≥ 2bc+ 2√

2bc · 2√bc = (2 + 2

√2)bc ≥ 4bc.

None of these results is correct, however, if an angle of ∆ABC is obtuse, as thefollowing example shows. Consider a triangle ABC with ∠BAC = 120◦ and

AB = AC. Then b = c, a = c√

3, and ap = bp = cp = a2 = c

√3

2 . One easily finds

that 3abc = 3c3√

3, while

a2(bp + cp) + b2(cp + ap) + c2(ap + bp) = 5c3√

3.

Also,

ap + bp + cp =3c√

3

2>

(2 +√

3)c

2= s.

4042. Proposed by Leonard Giugiuc and Diana Trailescu.

Let a, b and c be real numbers in [0, π/2] such that a + b + c = π. Prove theinequality

2√

2 sina

2sin

b

2sin

c

2≥√

cos a cos b cos c.

We received 14 correct solutions. We present the solution by Scott Brown. Similarsolutions came from Arslanagic Sefket, Michel Bataille, Andrea Fanchini, andJohn Heuvel.

In [1] and [2] respectively, we find the identities

sina

2sin

b

2sin

c

2=

r

4R(1)

and

cos a cos b cos c =s2 − 4R2 − 4Rr − r2

4R2, (2)

where R, r, and s are the circumradius, inradius, and semiperimeter of the triangle.We square both sides of the original inequality to obtain the equivalent statement

8 sin2 a

2sin2 b

2sin2 c

2≤ cos a cos b cos c,

Crux Mathematicorum, Vol. 42(5), May 2016

SOLUTIONS /227

into which we substitute the identities (1) and (2). The resulting inequality isequivalent to one due to Gerretsen [3]:

s2 ≤ 4R2 + 4Rr + 3r2.

References

[1] Anders Bager. “A family of goniometric inequalities.” Univ. Beograd. Publ.Elektrotehn. Fak. Ser. Mat. Fiz., No. 338 352 (1971), p. 10.

[2] Anders Bager. “Another family of goniometric inequalities.” Univ. Beograd.Publ. Elektrotehn. Fak. Ser. Mat. Fiz., No. 412 460 (1973), p. 209.

[3] D. S. Mitrinovic et al. Recent Advances in Geometric Inequalities. Kluwer,Dordrecht, 1989.

Editor’s Comment. Digby Smith pointed out that the inequality is equivalent toCrux Problem 974, proposed by Jack Garfunkel in Volume 10, (8), October 1984,and solved by Murray Klamkin in Volume 11 (10), December 1985. The solutionto 974 is based on Crux Problem 836, proposed by Vedula N. Murty in Volume9 (4), April 1983, and solved, again by Klamkin, in Volume 10 (7), August 1984.


Suppose that the lines m and n intersect at A and are not perpendicular. LetB be a point on n, with B 6= A. If F is a point of m, distinct from A, showthat there exists a unique conic CF with focus F and focal axis BF , intersectingn orthogonally at A. Given ε > 0, how many of the conics CF have eccentricity ε?

We recieved two correct solutions and present the solution submitted by the pro-poser.

Since m 6= n, the perpendicular to m through F and the perpendicular t to n atA intersect, say at K. Note that K is distinct from both F and A (since F 6= A).Define p to be the perpendicular to BF through K. Then A /∈ p (otherwise wewould have t ⊥ BF , implying n ‖ BF , a contradiction). We also have F /∈ p(otherwise KF ⊥ BF , implying BF = m and B ∈ m, contradicting B 6= A).

We first show uniqueness: Suppose that CF exists. Note that t is the tangentto CF at A. Since ∠KFA = 90◦ and K ∈ t, K must be on the directrix of CFassociated with F (see [2], Theorem 1 p. 14). Thus, CF must be the unique conicwith focus F , directrix p and eccentricity AF

d(A,p) . Conversely, because the line p

misses the distinct points A and F , we can consider the unique conic with focus F ,directrix p and eccentricity AF

d(A,p) . This conic passes through A (by the definition

of eccentricity) and is tangent to AK at A (since K ∈ p and ∠KFA = 90◦); ittherefore intersects n orthogonally at A. Also, its focal axis is BF (since BF ⊥ p).Thus, this conic satisfies the required conditions for CF .

Note that the eccentricity of CF is also equal to FBFA (see [2], Theorem 4, p. 18)

and that if F, F ′ are two distinct points on m (F, F ′ 6= A), then the conics CF and


228/ SOLUTIONS

CF ′ are distinct (their focal axes are distinct). From these remarks, we see thatthere are as many conics CF with eccentricity ε as points of M ∈ m that belong tothe locus E of points for which MB

MA = ε. If ε = 1, E is the perpendicular bisectorof AB; it intersects m (since m and n are not perpendicular), so that exactly oneconic CF is a parabola. If ε 6= 1, then E is a circle—the circle of Apollonius—whichcan intersect m in at most two points. The collection of all these circles (as ε variesover the positive real numbers except 1) forms a nonintersecting pencil of circleswith limiting points A and B, one through each point of the plane not on theperpendicular bisector of AB (see [1], Section 6.6). It follows that there are atmost two conics CF corresponding to a given value of ε.

To be more specific, E has diameter JJ ′ where J, J ′ are the points of n defined

by (1 + ε)−→AJ =

−−→AB and (1 − ε)

−−→AJ ′ =

−−→AB. The centre U of E is such that

(1 − ε2)−→AU =

−−→AB; its radius is ρ = JU = εAU = εAB

|1−ε2| . If H,H ′ are the

orthogonal projections of B,U onto m, respectively, then UH′

BH = AUAB = 1

|1−ε2| ;

hence UH ′ = BH|1−ε2| (where BH is the distance from B to m which, of course, is

always less than AB). We conclude that no, one, or two conics CF have eccentricityε according as UH ′ is greater than, equal to, or less than ρ, which is equivalent toε less than, equal to, or greater than BH

AB . So, for example, CF could never be acircle (for which ε = 0).

References

[1] H.S.M. Coxeter, Introduction to Geometry, Wiley, 1961.

[2] C. V. Durell, A Concise Geometrical Conics, MacMillan, 1952.

4044. Proposed by Dragoljub Milosevic.

Let x, y, z be positive real numbers such that x+ y + z = 1. Prove that

x+ 1

x3 + 1+

y + 1

y3 + 1+

z + 1

z3 + 1≤ 27

7.

We received 24 submissions, of which 22 were correct and complete. There weretwo main approaches: Jensen’s inequality, or comparing each term to a linearfunction. We present two solutions, one for each approach.

Solution 1, by Fernando Ballesta Yague.

As x + y + z = 1 and x, y, z are positive, we have x, y, z ∈ (0, 1). For x in theinterval (0, 1), consider the rational function

f(x) =x+ 1

x3 + 1=

1

x2 − x+ 1.


SOLUTIONS /229

Let’s take the second derivative to check its convexity:

f ′(x) =−2x+ 1

(x2 − x+ 1)2,

f ′′(x) =−2(x2 − x+ 1)2 − (−2x+ 1) · 2 · (x2 − x+ 1) · (2x− 1)

(x2 − x+ 1)4=

6x(x− 1)

(x2 − x+ 1)3.

Since x ∈ (0, 1), we have x − 1 < 0, but 6x > 0 and x2 − x + 1 > 0 (note that−x + 1 > 0 for x ∈ (0, 1)). So in the interval x ∈ (0, 1) we have f ′′(x) < 0 andhence f is concave. By Jensen’s Inequality for concave functions,

1

3(f(x) + f(y) + f(z)) ≤ f

Å1

3(x+ y + z)

ã= f

Å1

3

ã=

9

7;

in other words,

1

x2 − x+ 1+

1

y2 − y + 1+

1

z2 − z + 1≤ 27

7,

which is equivalent to the inequality we wanted to prove. Note that equality holdswhen x = y = z = 1

3 .

Solution 2, by Paul Bracken.

As in the previous solution, define f(x) = 1x2−x+1 and show that f is concave for

x ∈ (0, 1). Therefore, if t(x) is a tangent line to f(x) at some point x0 ∈ (0, 1)then the inequality f(x) ≤ t(x) holds for x ∈ (0, 1). Let us calculate the tangentline to f(x) at x0 = 1

3 :

t(x) = f ′Å

1

3

ã·Åx− 1

3

ã+ f

Å1

3

ã=

27

49x+

54

49.

The inequality f(x) ≤ t(x) then gives us x+1x3+1 ≤

2749x + 54

49 for x ∈ (0, 1). Weobtain similar inequalities by replacing x by y and z respectively, then add thethree inequalities to get

x+ 1

x3 + 1+

y + 1

y3 + 1+

z + 1

z3 + 1≤ 27

49(x+ y + z) + 3 · 54

49=

27

7,

where for the last equality we used x+ y + z = 1.

4045. Proposed by Galav Kapoor.

Suppose that we have a natural number n such that n ≥ 10. Show that by changingat most one digit of n, we can compose a number of the form x2 +y2 +10z2, wherex, y, z are integers.

We received two correct solutions. We present the solution by Roy Barbara.

Recall that Legendre’s three-square theorem states that a natural number is a sumof three squares if and only if it is not of the form 4m(8k + 7). In particular, anynatural number of the form 4k + 2 is a sum of three squares.


230/ SOLUTIONS

Now let 2k+1 be any odd natural number. Then we can write 4k+2 = a2+b2+c2.Using a2 + b2 + c2 ≡ 2 (mod 4), it is clear that exactly one of a, b, c is even, say c.Setting x = 1

2 (a+ b), y = 12 (a− b), c = 2z yields

4k + 2 = 2x2 + 2y2 + 4z2,

whence2k + 1 = x2 + y2 + 2z2.

Finally let n ≥ 10. By changing the last digit of n to a 5 (if necessary), we obtaina number of the form 10k + 5 for which we have

10k + 5 = 5x2 + 5y2 + 10z2 = (2x+ y)2 + (2y − x)2 + 10z2.


Let a, b, c be nonnegative real numbers such that√a+√b+√c ≥ 1. Prove that

a2 + b2 + c2 + 7(ab+ bc+ ca) ≥»

8(a+ b)(b+ c)(c+ a).

Two correct solutions were received. A purported counterexample that was submit-ted had an error. We present both solutions.

Solution 1, by Madhav R. Modak.»8(a+ b)(b+ c)(c+ a)

≤ (√a+√b+√c)»

8(a+ b)(b+ c)(c+ a)

=»

(4ab+ 4ca)[2(a+ b)(c+ a)] +»

(4bc+ 4ab)[2(a+ b)(b+ c)]

+»

(4ca+ 4bc)[2(b+ c)(c+ a)]

≤ 1

2[(4ab+ 4ca) + 2(a+ b)(c+ a)] +

1

2[(4bc+ 4ab) + 2(a+ b)(b+ c)]

+1

2[(4ca+ 4bc) + 2(b+ c)(c+ a)]

= 4(ab+ bc+ ca) + (a2 + ab+ ca+ bc) + (b2 + ab+ bc+ ca) + (c2 + ca+ bc+ ab)

= a2 + b2 + c2 + 7(ab+ bc+ ca),

which yields the desired result.


Since

(a2 + 3ab+ 3ca+ bc)2 = 8a(a+ b)(b+ c)(c+ a) + (a− b)2(a− c)2,


SOLUTIONS /231

it follows that

a2 + 3ab+ 3ca+ bc ≥√a(»

8(a+ b)(b+ c)(c+ a)).

Similarly

b2 + 3ab+ 3bc+ ca ≥√b(»

8(a+ b)(b+ c)(c+ a))

and

c2 + 3ca+ 3bc+ ab ≥√c(»

8(a+ b)(b+ c)(c+ a)).

Adding these three inequalities yields the result.

Editor’s comment. Equality holds if and only if a = b = c = 1/9.

4047?. Proposed by Abdilkadir Altintas.

Let ABC be a triangle with circumcircle O, orthocenter H and ∠BAC = 60◦.Suppose the circle with centre Q is tangent to BH, CH and the circumcircle ofABC. Show that OH ⊥ HQ.

All 14 submissions we received were correct. We feature two solutions.


232/ SOLUTIONS

Solution 1 is a composite of solutions by Vaclav Konecny and Edmund Swylan.

The statement of the problem is faulty: Because the plane is partitioned into asmany as eight regions by the circumcircle of triangle ABC and the lines HB andHC, there could be eight tritangent circles and, consequently, eight choices forQ, of which some lie on the Euler line OH (in which case the lines OH and HQwould be coincident, not perpendicular).

[Editor’s comment: Since the centres of all circles tangent to HB and HC wouldlie on a bisector of ∠BHC, the requirement that the circle be tangent also to thecircumcircle was perhaps included to limit the choice of tritangent circle to theincircle of the curvilinear triangle HBC (formed by the line segments HB andHC and the circular arc BC). Then the problem has been correctly stated for anacute ∆ABC, but it is still not correct when there is an obtuse angle at B or C.]

The exact location of Q is not relevant to the correct theorem:

For any circle tangent to the lines HB and HC, its centre Q mustbelong to one of the two bisectors of ∠BHC, and so must O.

The claim for Q is a familiar theorem, while the claim for O depends on ∠BAC =60◦ and must be proved.

As in the figure, denote the midpoints of AC and AB by M and N , respectively,and the feet of the altitudes to these lines by F and G. Then the segments FMand GN are congruent:

FM = |FA−MA| =∣∣∣∣AB · cos 60◦ − AC

2

∣∣∣∣ =1

2|AB −AC|

and

GN = |GA−NA| =∣∣∣∣AC · cos 60◦ − AB

2

∣∣∣∣ =1

2|AC −AB|.

Then the lines BF,CG,MO,NO form the sides of a rhombus for which the lineOH is a diagonal. Thus OH bisects one of the angles formed by the lines HF andHG, as claimed.

Solution 2 is a composite of similar solutions by Sefket Arslanagic, Ricardo BarrosoCampos, Prithwijit De (done independently), and Adnan Ibric with Salem Malikic.

As in the figure that accompanies the statement of the problem, we assume thatthe given triangle is acute, and that F and G are the feet of the altitudes fromB and C, respectively. Observe that ∠BHC = ∠FHG = 120◦ (since ∠A is60◦ and is opposite ∠FHG in the circle whose diameter is AH). Furthermore,∠BOC = 120◦ also (because O is the centre of the circumcircle so that the anglethere is twice the angle BAC = 60◦ which is inscribed in that circle). BecauseO and H are on the same side of BC, it follows that B,O,H,C are concyclic.Finally, note that because ∆BOC is isosceles, ∠OCB = 30◦. Since HQ is the


SOLUTIONS /233

bisector of ∠BHC,

∠BHQ =1

2∠BHC = 60◦.

Combine this with∠OHB = ∠OCB = 30◦,

and conclude that∠OHQ = ∠OHB + ∠BHQ = 90◦.

Editor’s Comments. Essentially the same problem has appeared before in Crux[1988: 165; 1990: 103] as Problem M1046, which was taken from the 1987 U.S.S.Rjournal Kvant:

If ∠A = 60◦ then one of the bisectors of the angle between the altitudesfrom B and C passes through O.

This and related properties were discussed under the heading “Property 3” inthe article “Recurring Crux Configurations 3: Triangles Whose Angles Satisfy2B = C +A” [2011: 350].


Let n ≥ 2 be an integer and let ak ≥ 1 be real numbers, 1 ≤ k ≤ n. Prove theinequality

a1a2 · · · an −1

a1a2 · · · an≥Åa1 −

1

a1

ã+

Åa2 −

1

a2

ã+ · · ·+

Åan −

1

an

ãand study equality cases.

Thirteen solutions were received, all of which established the inequality. Two ofthem did not get all the possible conditions for equality, while three others neglectedto consider when equality occurred. The solutions were all similar to the onepresented below.

Let

f(x) = x− 1

x

and observe that, for x, y ≥ 1,

f(xy)− f(x)− f(y) = (xy)−1(xy − 1)(x− 1)(y − 1) ≥ 0

with equality if and only if at least one of x and y is equal to 1.

We establish the result by induction.

The foregoing shows that it is true for n = 2. Suppose that the inequality holdsfor n = m ≥ 2 with equality iff all but at most one of a1, a2, . . . , am is equal to 1.Then, by the foregoing property of f and the result for n = m,

f(a1a2 · · · amam+1) ≥ f(a1a2 · · · am)+f(am+1) ≥m∑k=1

f(ak)+f(am+1) =m+1∑k=1

f(ak).


234/ SOLUTIONS

Equality holds if and only if either

• a1a2 · · · am = 1, in which case a1 = a2 = · · · = am = 1, or

• am+1 = 1 and f(a1a2 · · · am) =∑mk=1 f(ak).

In either case, all but at most one of a1, a2, . . . , am+1 is equal to 1.

Editor’s comments. One can also peel off the last two terms in the product so thatthe induction step becomes

f(a1a2 . . . amam+1) ≥m−1∑k=1

f(ak) + f(amam+1) ≥m+1∑k=1

f(ak).

Edmund Swyland observed that if, for any i and j, you replaced the pair (ai, aj)by (aiaj , 1), the left side f(a1a2 · · · an) of the inequality remained unchanged, butthe right side increased. Thus we can reduce the problem to establishing that itholds when all but two of the ai are equal to 1, and this now involves dealing withthe case n = 2.

Kee-Wai Lau pointed out that an easy induction argument yields

f(a1a2 · · · an)−n∑k=1

f(ak) =n∑k=2

(a1a2 · · · ak−1 − 1)(ak − 1)(a1a2 · · · ak − 1)

a1a2 · · · ak.


Evaluate ∫sinx− x cosx

(x+ sinx)(x+ 2 sinx)dx

for all x ∈ (0, π/2).

We received 16 submissions all of which were correct. We present a composite ofthe nearly identical solutions given by Adnan Ali, Michel Bataille, Prithwijit De,Joseph Ling and Albert Stadler, all done independently

Let I denote the given integral. Since it is readily checked that

(1 + cosx)(x+ 2 sinx)− (1 + 2 cosx)(x+ sinx) = sinx− x cosx,

we have

I =

∫ Å1 + cosx

x+ sinx− 1 + 2 cosx

x+ 2 sinx

ãdx

= ln(x+ sinx)− ln(x+ 2 sinx) + C

= ln

Åx+ sinx

x+ 2 sinx

ã+ C,

where C is an arbitrary constant.


SOLUTIONS /235

4050. Proposed by Mehtaab Sawhney.

Prove that2n∑k=0

Ç4n

k, k, 2n− k, 2n− k

å=

Ç4n

2n

å2

for all nonnegative integers n.

We received twelve correct solutions which were split between an arithmetic proofand a proof by double counting, so we present a solution of each type.

Solution 1, by C.R. Pranesachar.

We have

2n∑k=0

Ç4n

k, k, 2n− k, 2n− k

å=

2n∑k=0

Ç4n

2n

åÇ2n

k

å2

=

Ç4n

2n

å 2n∑k=0

Ç2n

k

åÇ2n

2n− k

å=

Ç4n

2n

å2

,

where the last equality is due to Vandermonde’s identity.

Solution 2, by Joseph DiMuro.

Let’s say we have a classroom with 4n students. The teacher wants to choose2n of them to work on one project and 2n of them to work on a second project(independently of each other – students may be assigned to both or neither of theprojects). In how many ways can the teacher assign students to the projects?

On one hand there are(4n2n

)ways to choose the students for each of the two projects,

thus(4n2n

)2possibilities altogether.

On the other hand note that if k students are assigned to both projects then 2n−kwill be assigned to just the first project, 2n − k to just the second project, andk to neither project. So the teacher can proceed as follows: first decide on thenumber k of students that will be assigned to both projects, then partition theclass into four groups – of size k (both projects), 2n − k (first project), 2n − k(second project), and k (neither project). There are

2n∑k=0

Ç4n

k, k, 2n− k, 2n− k

åways to do this. Thus the two sides in the problem are equal.


SOLUTIONS /271

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

Statements of the problems in this section originally appear in 2015 : 41(6), p. 260–263.


Let a, b and c be the side lengths of a triangle. Prove that

(a+ b+ c)(a2b2 + b2c2 + c2a2

)≥ 3abc

(a2 + b2 + c2

).

We received eleven correct solutions. We present two solutions.


A key to a solution is contained in solution 2 to Crux problem 3991 published in41 (1) (December 2015).

Setting a = y+z2 , b = z+x

2 , c = x+y2 transforms the proposed inequality into

x5+y5+z5+x2y2z+x2yz2+xy2z2 ≥ x3y2+x2y3+y3z2+y2x3+z3x2+z2x3, (1)

where x, y, z are positive real numbers. The general Schur inequality is

ur(u− v)(u− w) + vr(v − w)(v − u) + wr(w − u)(w − v) ≥ 0

for u, v, w ≥ 0 and r real. We take r = 12 and u = x2, v = y2, w = z2 and obtain

x(x2 − y2)(x2 − z2) + y(y2 − z2)(y2 − x2) + z(z2 − x2)(z2 − y2) ≥ 0.

Expanding and arranging directly leads to (1).


By Consequence 16.3, p. 156 from [1], all symmetric three-variable polynomials ofdegre less than or equal to five achieve their maximum and minimum values onR∗ at (a, b, c) if and only if (a− b)(b− c)(c− a) = 0 or abc = 0. It thus suffices toprove the given inequality for b = c and c = 0.

If b = c, then we have to prove that

(a+ 2b)(2a2b2 + b4) ≥ 3ab2(a2 + 2b2)

2a3b2 + ab4 + 4a2b3 + 2b5 ≥ 3a3b2 + 6ab4

4a2b3 + 2b5 ≥ a3b2 + 5ab4

b2(a− b)2(2b− a) ≥ 0,


272/ SOLUTIONS

which is true by the triangle inequality.

If c = 0, then we have to prove that

(a+ b)a2b2 ≥ 0,

which is true.

Equality holds if and only if a = b = c.

[1] Z. Cvetkovski, Inequalities - Theorems, Techniques and Selected Problems,Springer-Verlag, 2012.


Let k < 0 be a fixed real number. Let a, b, c and d be real numbers such thata+ b+ c+ d = 0 and ab+ bc+ cd+ da+ ac+ bd = k. Prove that abcd ≥ −k2/12and determine when equality holds.

We received six submissions, of which five were correct and complete. We presentthe solution by Oliver Geupel, slightly modified by the editor.

It is sufficient to consider the case when three of the numbers a, b, c, d have thesame sign. Otherwise, abcd ≥ 0 > −k2/12, and we’re done.

By symmetry, there is no loss of generality in assuming that it is a, b, and c whichhave the same sign. Then a+ b+ c+ d = 0 gives us d = −(a+ b+ c), so

k = ab+ bc+ cd+ da+ ac+ bd

= (a+ b+ c)d+ (ab+ bc+ ca)

= −(a+ b+ c)2 + (ab+ bc+ ca)

= −(a2 + b2 + c2 + ab+ bc+ ca),

and the inequality we want to prove can be rewritten as

−abc(a+ b+ c) ≥ −(a2 + b2 + c2 + ab+ bc+ ca)2/12,

or, equivalently,

12abc(a+ b+ c) ≤ (a2 + b2 + c2 + ab+ bc+ ca)2. (1)

Let x = |a|, y = |b|, z = |c|, so that x, y, z are nonnegative real numbers. Since a,b and c have the same sign by assumption,

abc(a+ b+ c) = xyz(x+ y + z), and

a2 + b2 + c2 + ab+ bc+ ca = x2 + y2 + z2 + xy + yz + zx,

so it is sufficient to prove that (1) holds with a,b and c replaced by x, y and z,respectively.

Crux Mathematicorum, Vol. 42(6), June 2016

SOLUTIONS /273

By the AM-GM Inequality (where the right hand side is treated as a sum of 8terms, with terms repeated as indicated by the coefficients), we have

8x2yz ≤ x4 + 2x3y + 2x3z + 3y2z2. (2)

The equality holds only when x4 = x3y = x3z = y2z2, that is when either x =y = z or x = yz = 0. Summing up inequality (2) and its two cyclic variants, andadding terms to both sides so we can complete the square on the right hand side,we obtain

12xyz(x+ y + z) ≤ (x2 + y2 + z2 + xy + yz + zx)2, (3)

where the equality holds if and only if x = y = z. This shows that (1) holds, andthus concludes the proof that abcd ≥ −k2/12.

It follows from the preceding steps that the equality holds if and only if three ofthe four numbers a, b, c, and d are equal. A straightforward computation (froma+ b+ c+ d = 0 and abcd = −k2/12) shows that the common value is ±

√−k/6,

and that the fourth number has the value ∓3√−k/6.

4053. Proposed by Sefket Arslanagic.

Prove thatcosα cosβ

cos γ+

cosβ cos γ

cosα+

cosα cos γ

cosβ≥ 3

2,

where α, β and γ are angles of an acute triangle.

We received 13 correct solutions. We present a composite of essentially the samesolution by Jose Luis Dıaz-Barrero, Dionne Bailey, Elsie Campbell, and CharlesR. Diminnie (joint), Henry Ricardo, and Lorian Saceanu.

Since α, β, γ ∈ (0, π2 ) we have

cosα cosβ

cos γ=

cosα cosβ

cos γ· tanα+ tanβ

tanα+ tanβ

=sinα cosβ + cosα sinβ

cos γ· 1

tanα+ tanβ

=sin(π − γ)

cos γ· 1

tanα+ tanβ

=tan γ

tanα+ tanβ.

By the well-known Nesbitt’s Inequality which states that

a

b+ c+

b

c+ a+

c

a+ b≥ 3

2

with equality if and only if a = b = c, we then have∑cyc

cosα cosβ

cos γ=∑cyc

tan γ

tanα+ tanβ≥ 3

2


274/ SOLUTIONS

with equality if and only if α = β = γ ; in other words, if the given triangle isequilateral.

Editor’s comment. Both Ricardo and Diminnie et al pointed out that the currentproblem is the special case when m = 0 of problem #5381 in the January 2016issue of School, Science and Mathematics :

If A, B, C are angles of an acute triangle, then∑ÅcosA cosB

cosC

ãm+1

≥ 3

2m+1

for all nonnegative integers m.

Interestingly, this general problem was proposed by D. M. Batinetu-Giurgiu andNeculai Stanciu, two regular contributors to the Crux problem section. Ricardoactually gave a proof to the general inequality as well as a proof for the above case.The general proof uses the same argument in the featured solution above togetherwith the power mean inequality and it appeared as solution 3 in the April 2016issue of that journal (see www.ssma.org/publications).


Find a prime p such that the number

(p2 − 4)2 − 117(p2 − 4) + 990

has a minimum digit sum.

We received eight correct and complete solutions, all of which were very similar.We present the solution by Joseph DiMuro.

Letf(p) = (p2 − 4)2 − 117(p2 − 4) + 990.

If p 6= 3 is a prime, then p = 3n± 1 for some integer n. Then

p2 − 4 = 9n2 ± 6n− 3,

a multiple of 3 ; thus, (p2−4)2 is a multiple of 9. But 117 and 990 are also multiplesof 9. So f(p) is a multiple of 9, as is its digit sum.

If we had f(p) = 0, then by the quadratic formula we would have

p2 − 4 =117±

√9729

2,

which has no integer solutions, so f(p) 6= 0. Therefore, the digit sum of f(p) mustbe at least 9 when p is a prime other than 3.

However, the digit sum of f(3) = 430 is 7, so the minimum digit sum is 7, attainedfor p = 3.


SOLUTIONS /275


Prove that if x, y > 0, x 6= y and 0 < a < b < 12 < c < d < 1 then :

x[(yx

)a+(yx

)d−(yx

)b−(yx

)c]> y[(xy

)b+(xy

)c−(xy

)a−(xy

)d].

We received three correct solutions and feature two of them that are similar.


With t = xy , the inequality can be re-written as

(td + t1−d)− (tc + t1−c) > (tb + t1−b)− (ta + t1−a). (1)

Let us fix t > 0, t 6= 1 and set f(u) = tu and g(u) = f(u) + f(1− u) so that (1) isjust

g(d)− g(c) > g(b)− g(a). (2)

From the Mean Value Theorem, we have

g(b)− g(a) = (b− a)g′(α), g(d)− g(c) = (d− c)g′(β)

for some α ∈ (a, b) and β ∈ (c, d).

Since g′(u) = (ln t)(tu − t1−u) = (ln t)(f(u)− f(1− u)), (2) becomes

(d− c)(ln t)(f(β)− f(1− β)) > (b− a)(ln t)(f(α)− f(1− α)). (3)

Applying the Mean Value Theorem again, we have f(β)−f(1−β) = (2β−1)(ln t)tσ

and f(α)− f(1−α) = (2α− 1)(ln t)tτ with σ between β and 1− β and τ betweenα and 1− α.

Substituting into (3) and because (ln t)2 > 0, we are reduced to proving

(d− c)(2β − 1)tσ > (b− a)(2α− 1)tτ . (4)

Now, on the one hand d− c > 0, tσ > 0, 2β − 1 > 0 (note that β ∈ (c, d), henceβ > 1

2 ) and on the other hand, b − a > 0, tτ > 0, 2α − 1 < 0 (since α ∈ (a, b)).Thus,

(d− c)(2β − 1)tσ > 0 > (b− a)(2α− 1)tτ

and (4) follows.

Solution 2, by Daniel Sitaru and Leonard Giugiuc.

Let f : [0, 1]→ R, f(α) = x1−αyα+xαy1−α

2 , with x, y ∈ (0,∞), x 6= y. We have :

limα→0α>0

f(α) =x+ y

2, lim

α→1α<1

f(α) =x+ y

2.


276/ SOLUTIONS

Since

f ′(α) =1

2(ln y − lnx)(x1−αyα − xαy1−α),

then

f ′(α) = 0 ⇒ x1−αyα = xαy1−α ⇒(xy

)1−2α= 0⇒ a =

1

2.

Therefore, min f(α) = f(

12

)=√xy and f(a) > f(b) > f

(12

), f

(12

)< f(c) <

f(d). By adding, we get f(a) + f(d) > f(b) + f(c) or

x1−aya + xay1−a

2+x1−dyd + xdy1−d

2>x1−byb + xby1−b

2

>x1−byb + xby1−b

2+x1−cyc + xcy1−c

2.

This results in

x[(yx

)a+(yx

)d−(yx

)b−(yx

)c]> y[(xy

)b+(xy

)c−(xy

)a−(xy

)d].

4056. Proposed by Idrissi Abdelkrim-Amine.

Let n be an integer, n ≥ 2. Consider real numbers ak, 1 ≤ k ≤ n such thata1 ≥ 1 ≥ a2 ≥ . . . ≥ an > 0 and a1a2 . . . an = 1. Prove that

n∑k=1

ak ≥n∑k=1

1

ak.

We received eight solutions of which six were correct. We present the solution byRoy Barbara.

Note first that if 0 < a, b ≤ 1, then

a− 1

a+ b− 1

b≥ ab− 1

ab. (1)

Indeed, multiplying by ab, (1) is equivalent, in succession, to

a2b− b+ ab2 − a ≥ (ab)2 − 1or (a+ b)(ab− 1) ≥ (ab+ 1)(ab− 1)or (1− ab)(1− a)(1− b) ≥ 0,

which is true.

We now prove the given inequality by using induction on n ≥ 2.


SOLUTIONS /277

The case when n = 2 is trivial. Suppose the inequality holds for some n ≥ 2,and let ai, i = 1, 2, · · · , n + 1 satisfy a1 ≥ 1 ≥ a2 ≥ · · · ≥ an ≥ an+1 anda1a2 · · · anan+1 = 1. We need to prove that(

n−1∑k=1

ak −n−1∑k=1

1

ak

)+

Åan −

1

an

ã+

Åan+1 −

1

an+1

ã≥ 0. (2)

Set bn = anan+1.

Then clearly a1 ≥ 1 ≥ a2 ≥ · · · ≥ an−1 ≥ bn and a1a2 · · · an−1bn = 1.

By the induction hypothesis, we have(n−1∑k=1

ak −n−1∑k=1

1

ak

)+

Åbn −

1

bn

ã≥ 0. (3)

Hence, to get (2), it suffices to prove thatÅan −

1

an

ã+

Åan+1 −

1

an+1

ã≥Åbn −

1

bn

ã,

that is, Åan −

1

an

ã+

Åan+1 −

1

an+1

ã≥Åanan+1 −

1

anan+1

ã,

which is true by (1) as 0 < an, an+1 ≤ 1.

Editor’s comment. The proposer of the current problem remarked that the gi-ven inequality was inspired by the following inequality due to Leonard Giugiuc :(∑ak)2 ≥ n

∑ 1ak

, where the ak’s satisfy the same conditions given in the currentproblem.

4057. Proposed by Eeshan Banerjee.

Let ABC be a non-obtuse triangle with circumradius R, inradius r and area ∆.Prove that

∆ <

Ç1r + 3R+ 3

7

å7

.

We received four correct solutions. We present a composite of very similar solutionsby Michel Bataille and Andrea Fanchini.

With AM-GM, we have that

1r +R+R+R+ 1 + 1 + 1

7≥ 7

…R3

r,

so it suffices to prove that

∆ <R3

r.


278/ SOLUTIONS

From Euler’s inequality R ≥ 2r and the inequality R ≥ 2s3√3, we have

R3

r≥ 4r2 · 2sr · 3√

3=

8√

3

9· rs =

8√

3

9·∆ > ∆,

completing the proof.

4058. Proposed by Francisco Javier Garcıa Capitan.

Let ABC be a triangle. For any X on line BC, let Xb and Xc be the circumcentersof the triangles ABX and AXC, and let P be the intersection point of BXc

and CXb. Prove that the locus of P as X varies along the line BC is the conicthrough the centroid, orthocenter, and vertices B and C, and whose tangents atthese vertices are the corresponding symmedians. (Recall that a symmedian is thereflection of a median in the bisector of the corresponding angle.)

We received two submissions, both of which were correct. We feature the solutionby Michel Bataille with a few details added from the proposer’s solution.

As usual, set BC = a,CA = b, AB = c. Should a = b, then CXb will be theperpendicular bisector of AB, which immediately implies that this line is thelocus of P ; similarly, should a = c, then the locus of P would be the perpendicularbisector of AC. Assume therefore that a 6= b, c. We shall see that under this furtherassumption the locus is a hyperbola. For our argument we shall use barycentriccoordinates relative to (A,B,C), and the following notation :

SA =b2 + c2 − a2

2, SB =

c2 + a2 − b2

2, SC =

a2 + b2 − c2

2.

For later use, here are a few readily checked relations satisfied by these numbers :

SB + SC = a2, SC + SA = b2, SA + SB = c2,

and

a2SA + SBSC = b2SB + SCSA = c2SC + SASB

= SASB + SBSC + SCSA =1

2(a2SA + b2SB + c2SC).

If (f : g : h) is the point at infinity of a line `, then (gSB − hSC : hSC − fSA :fSA − gSB) is the point at infinity of the perpendiculars to `. With the help ofthis property, we easily obtain the equations of the perpendicular bisector `1 ofAB and `2 of AC,

`1 : c2x− c2y + (a2 − b2)z = 0, `2 : b2x+ (a2 − c2)y − b2z = 0.

If X = (0 : β : γ) with β+γ = 1, the point at infinity on AX is (−1 : β : γ), hencethe one on the perpendiculars to AX is (βSB − γSC : γSC + SA : −SA − βSB). Itfollows that the perpendicular bisector m of AX is

x(β2SB + γ2SC + SA) + y(γ2a2 − c2) + z(β2a2 − b2) = 0.


SOLUTIONS /279

Note that

β2SB + γ2SC + SA

= β2SB + (1− β)2SC + SA

= a2β2 − 2βSC + b2 = a2β2 − β(a2 + b2 − c2) + b2 = βc2 + γb2 − a2βγ.

Let x1 = a2(SA + βSB), y1 = a2βSA + b2SB , z1 = c2(SC − a2β). It is ratherlong but easy to check that (x1, y1, z1) satisfies both the equations of `1 and m.Thus Xb = (x1 : y1 : z1). Similarly, we obtain Xc = (x2 : y2 : z2) with x2 =a2(SA + γSC), y2 = b2(SB − a2γ), z2 = a2γSA + c2SC .

The equations of CXb : xy1 − yx1 = 0 and BXc : xz2 − zx2 = 0 then provideP = (u : v : w) = (x1x2 : x2y1 : x1z2) so that

v =a2βSA + b2SBa2(SA + βSB)

· u, w =a2γSA + c2SCa2(SA + γSC)

· u,

from which we obtain

a2β =b2SBu− a2SAvvSB − uSA

, a2γ =c2SCu− a2SAwwSC − uSA

.

Eliminating β, γ (through β + γ = 1) yields a necessary and sufficient conditionon u, v, w for P to belong to the desired locus, namely

u2SA(a2SA + b2SB + c2SC)− uvc2(a2SA + SBSC)− wub2(a2SA + SBSC)

+vwa2(SASB + SBSC + SCSA) = 0;

that is, 2u2SA − c2uv − b2uw + a2vw = 0.

Thus, the locus of P is the conic Γ with equation

(b2 + c2 − a2)x2 − c2xy + a2yz − b2zx = 0.

Note that because we have assumed that a 6= b, c, the discriminant of the conic

(namely, a2

4 (b2 − a2)(c2 − a2)) is nonzero ; because the coefficient of y2 is zero,this nondegenerate conic must be a hyperbola, as claimed. Let C(x, y, z) be theleft-hand side of the equation. We readily find that

C(0, 1, 0) = C(0, 0, 1) = C(1, 1, 1) = C(SBSC , SCSA, SASB) = 0,

hence Γ passes through B,C,G,H (respectively), where G = (1 : 1 : 1) andH = (SBSC : SCSA : SASB) denote the centroid and the orthocenter of ABC. Inaddition, the equation of the tangent to Γ at (x0 : y0 : z0) is

2xx0SA −1

2c2(x0y + xy0)− 1

2b2(x0z + xz0) +

1

2a2(y0z + yz0) = 0.

In particular, the tangent to Γ at B is c2x − a2z = 0, a line passing through theLemoine point K = (a2 : b2 : c2) and through B. Therefore this tangent is the


280/ SOLUTIONS

symmedian through vertex B. Similarly, the tangent to Γ at C is the symmedianthrough vertex C.

Editor’s comments. The proposer noted the following theorem to be a consequenceof his problem :

If a conic passes through the vertices B and C of a triangle ABC,while the tangents at those points are the corresponding symmedians,then the centroid of the triangle lies on the conic if and only if theorthocenter does also.

4059. Proposed by Marcel Chirita.

Let a, b ∈ (0,∞), a 6= b. Determine the functions f : R 7→ R \ {0} such that

f(ax) = exf(bx), ∀x ∈ R.

We received four submissions of which three were correct and complete. We presentthe solution by Michel Bataille.

We show the following :

Let P be the set of all functions from R to R \ {0} that are periodicwith period ln(b/a) and let u : t 7→ u(t) = ln(|t|) for t 6= 0. Then

the solutions are the functions t 7→ g(t) · et

a−b where the function g isdefined by

g(t) = p(u(t)) (t > 0), g(0) = α, g(t) = q(u(t)) (t < 0)

for some p, q ∈ P and some α ∈ R \ {0}.

First, a remark : if we set g(x) = f(x) · e−xa−b , a simple calculation shows that

solving the given equation boils down to solving the functional equation g(ax) =g(bx) for functions g : R→ R \ {0}, f being then defined by f(x) = g(x) · e

xa−b for

x ∈ R. Since xa takes all real values when x does, substituting x

a for x even reducesthe problem to seeking functions g : R→ R \ {0} such that

g(x) = g

Åx · b

a

ã(1)

for all x ∈ R. Let g be a solution. Then, for x > 0, we have

g(eln x) = g(eln x+ln(b/a)),

that is, p(lnx) = p(lnx + ln(b/a)) if we set p = g ◦ exp. Since lnx takes all realvalues as x describes (0,∞), it follows that p is periodic with period ln(b/a) anddoes not take the value 0, i.e. p ∈ P, and that g(x) = p(lnx) = p(ln(|x|) forpositive x.

Similarly, for x < 0, we have

g(−eln(−x)) = g(−eln(−x)+ln(b/a))


SOLUTIONS /281

that is, q(ln(−x)) = q(ln(−x) + ln(b/a)) where q = g ◦ (− exp) is an element of P,so that g(x) = q(ln(−x)) = q(ln(|x|).

Conversely, define g by

g(t) = p(u(t)) (t > 0), g(0) = α, g(t) = q(u(t)) (t < 0)

where p, q ∈ P and α ∈ R \ {0}. Then, g takes only nonzero values and, if x > 0,g(x ·(ba

))= p

(ln(x ·(ba

)))= p(lnx + ln(b/a)) = p(lnx) = g(x) and if x < 0,

then g(x ·(ba

))= q

(ln(−x ·

(ba

)))= q(ln(−x) + ln(b/a)) = q(ln(−x)) = g(x).

Thus, the equality g(x ·(ba

))= g(x) holds for all real x (it is obvious for x = 0)

and so g satisfies (1) with g(x) 6= 0 for all real x. The proof is complete.

Remark. The solutions which are continuous at 0 are the functions t 7→ αet

a−b

where α is a nonzero real constant : with the above notations, f is continuous at0 if and only if g is. So, we consider a solution g of (1), with g continuous at 0.Suppose first that b < a. Then,

g

Çx ·Åb

a

ãkå= g

Çx ·Åb

a

ãk+1å

if k is a positive integer ; hence, by an immediate induction, we see that g(x) =

gÄx ·(ba

)näfor all positive integers n. Since 0 < b

a < 1, we have limn→∞

x ·(ba

)n= 0

and so limn→∞

gÄx ·(ba

)nä= g(0). As a result, g(x) = g(0) for any x ∈ R. If b > a,

the treatment is similar using the equation g(x) = g(x · ab

)which holds for all x as

well. Thus g is a constant function. Conversely, any constant function R→ R\{0}is obviously a solution of (1).

Editor’s comments. Roy Barbara proposed and solved the following generalizationof the problem. Set k = a/b and t = bx (t ∈ R). If ϕ : R→ R∗ is a function withϕ(0) = 1 and k ∈ (0,∞), k 6= 1, determine all the functions f : R→ R∗ satisfying

f(kt) = ϕ(t) · f(t) ∀t ∈ R.

Unfortunately, Marcel Chirita passed away on 29 February 2016 and he cannotenjoy the beautiful solution given above. We will miss him and we will miss hisprecious contribution to the journal.


Let

f(x, y) =xy(x+ y)

(1− x− y)3.

Find the range of f(x, y) when its domain is restricted to the circle S that satisfiesthe equation x2 + y2 = 1− 2x− 2y.

Five correct solutions were submitted from four people. Two others made incorrectsubmissions. We present two solutions.


282/ SOLUTIONS

Solution 1, following the approach of Kee-Wai Lau.

Suppose that (x, y) ∈ S and let t = x+ y. Then

(t+ 2)2 = 2(x2 + y2 + 2x+ 2y − 1) + 6− (x− y)2 = 6− (x− y)2 ≤ 6,

so that −2−√

6 ≤ t ≤ −2 +√

6 < 1, with equality possible only if x = y. Since

2xy = (x+ y)2 + 2(x+ y)− 1 = t2 + 2t− 1,

we have

f(x, y) =t(t2 + 2t− 1)

2(1− t)3.

The condition for S can be written as

(x+ 1)2 + (y + 1)2 = 3,

so that S is a circle with centre (−1,−1) that intersects the line y = x atÅ1

2(−2−

√6),

1

2(−2−

√6)

ãand

Å1

2(−2 +

√6),

1

2(−2 +

√6)

ã.

These points correspond to the limiting values of t. From the geometry, it is easilyseen that as (x, y) ranges over S, the variable t assumes all values in the closedinterval [−2−

√6,−2 +

√6]. [Alternatively, using the theory of the quadratic, one

can determine that the system®(x+ 1) + (y + 1) = t+ 2,

(x+ 1)2 + (y + 1)2 = 3,

is solvable for real values of x and y if and only if (t+ 2)2 ≤ 6.]

Since f(x, y) = −√

6/18 when t = −2 −√

6 and f(x, y) =√

6/18 when t =−2 +

√6, and since f is continuous in t, the range of f includes the closed interval

[−√

6/18,√

6/18].

Observe that

f(x, y) +

√6

18=

(9−√

6)t3 + (18 + 3√

6)t2 − (9 + 3√

6)t+√

6

18(1− t)3

=(t+ 2 +

√6)[(9−

√6)t2 + (6− 4

√6)t+ (3−

√6]

18(1− t)3.

Since the quadratic factor, having zero discriminant, is the square of a linearpolynomial, and since 1− t > 0, f(x, y) ≥ −

√6/18 when t ≥ −2−

√6. Likewise

f(x, y)−√

6

18=

(9 +√

6)t3 + (18− 3√

6)t3 + (−9 + 3√

6)t−√

6

18(1− t)3

=(t+ 2−

√6)[(9 +

√6)t2 + (6 + 4

√6)t+ (3 +

√6)]

18(1− t)3≤ 0,


SOLUTIONS /283

so that f(x, y) ≤√

6/18 when t ≤ −2 +√

6. Thus the range of f is exactly[−√

6/18,√

6/18].


Let u = x2 + y2 and

a =2x

u+ 1b =

2y

u+ 1c =

u− 1

u+ 1.

When (x, y) ∈ S, we have that

u+ 1 = 2(1− x− y) and u− 1 = x2 + y2 − 1 = −2(x+ y).

It can be checked that

a+ b+ c = 0, a2 + b2 + c2 = 1 and so ab+ bc+ ca = −1

2.

Also

abc = − xy(x+ y)

(1− x− y)3= −f(x, y).

Thus, the real numbers a, b, c are the roots of the polynomial

t3 − 1

2x+ f(x, y).

Since the roots are all real, we must have 4 · (1/8) ≥ 27(f(x, y))2, so that

− 1√54≤ f(x, y) ≤ 1√

54

and f(x, y) belongs to the closed interval [−√

6/18,√

6/18].

Conversely, let p ∈ [−√

6/18,√

6/18]. We show that p = f(x, y) for some (x, y) ∈S. This is true for p = 0, since f(0,−1 +

√2) = 0. Suppose p 6= 0. Consider

the polynomial t3 − 12 t + p and let a, b, c be its roots. Since the discriminant

condition 4 · (1/8) ≥ 27p2 holds, the roots are all real. Since a + b + c = 0 andab+ bc+ ca = −1/2, we have that a2 + b2 + c2 = 1. Since abc 6= 0, −1 < c < 1, sothat c = (v − 1)/(v + 1) for some v > 0. Now, let

x =a(1 + v)

2and y =

b(1 + v)

2.

Then x2 + y2 = v = 1− 2x− 2y, so that (x, y) ∈ S. Moreover

f(x, y) =xy(x+ y)

(1− x− y)3=ab

4(v + 1)2 · (v + 1)(a+ b)

2· 8

(v + 1)3

= ab(a+ b) = −abc = p.


284/ SOLUTIONS

Editor’s Comments. The proposer submitted a second solution that followed thesame strategy as Lau, except that he analyzed the behaviour of the function t(t2 +2t− 1)(1− t)−3 by calculus. Paul Bracken used Lagrange Multipliers and locatedthe maximum and minimum values of f(x, y) as well as eight other critical pointson S given by the equations (x+1)2+(y+1)2 = 3 and (x2+y2)(x+y) = xy+x+y.

Paul Deiermann looked at the more general restriction (x + 1)2 + (y + 1)2 = awhere 0 < a < 9/2, and parameterized the points of S by

(x, y) = (−1 +√a cos θ,−1 +

√a sin θ).

He found f(x, y) to be equal to

1

2· [q2 − 2q + 2− a][−2 + q]

(3− q)3,

where q =√

2a cos(θ − π/4) satisfies −√

2a ≤ q ≤√

2a. He then analyzed thisfunction, identifying its values at the endpoints and the two critical points of theinterval [−

√2a,√

2a]. In the case a = 3 of the problem, the global maximumis achieved at both an endpoint and a critical point, while the global minimumis achieved at the other endpoint and critical point. He adds that Mathematicagraphs suggest that a = 3 is the only value of a where f achieves both globalextrema at an endpoint and a critical point at the same time.


318/ SOLUTIONS

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



Let ABC be a non-obtuse triangle none of whose angles are less than π4 . Find the

minimum value of sinA sinB sinC.

We received twelve submissions, of which 10 were correct and one was faulty. Wepresent two solutions.

Solution 1, by Adnan Ali.

Let us first fix angle A and determine the values of B and C for which the productsinA sinB sinC is smallest. Since A is fixed, this product is minimized if and onlyif

sinB sinC =cos(B − C)− cos(B + C)

2=

cos(B − C)− cos(π −A)

2is minimized. But cos(π−A) is fixed and so it is enough to minimize cos(B −C).Because the triangle is not obtuse and all angles are not less than π/4, we have|B − C| ≤ π/4. Since the cosine function is decreasing over [0, π/2], cos(B −C) is minimized if |B − C| is maximum, and that happens when B = π/4 andC = 3π/4− A, or vice-versa. So now we have reduced the problem of finding theminimum value of the given product to finding the minimum value of

sinA sin(π/4) sin(3π/4−A), π/4 ≤ A ≤ π/2.

This is quickly done by minimizing sinA sin(3π/4 − A) = cos(3π/4−2A)−cos(3π/4)2 ,

which is same as minimizing cos(3π/4−2A), where π/4 ≤ A ≤ π/2. The bounds onA imply that −π/4 ≤ 3π/4− 2A ≤ π/4, and so the minimum value of cos(3π/4−2A) is achieved for 3π/4 − 2A = −π/4 or π/4 ; each of these values leads toan isosceles right triangle. Thus, the minimum value of sinA sinB sinC is 1/2,achieved for an isosceles right triangle ABC.

Solution 2, by Daniel Dan.

We use the identity sinA sinB sinC = 14 (sin 2A+ sin 2B + sin 2C). Define

f(x) :[π

2, π]→ [0, 1], f(x) = sinx,

and note that the function is concave ; in particular, every point of the graph off(x) except for its end points, namely

(π2 , 1)

and (π, 0), lies above the line

g(x) = − 2

πx+ 2

Crux Mathematicorum, Vol. 42(7), September 2016

SOLUTIONS /319

that joins those end points. Consequently, we have

1

4((f(2A) + f(2B) + f(2C)) ≥ 1

4((g(2A) + g(2B) + g(2C))

=1

4

Å−2(2A+ 2B + 2C)

π+ 6

ã=

1

2.

We conclude that the product sinA sinB sinC cannot be less than 12 when all

three angles are restricted to the domain[π4 ,

π2

]. The minimum is achieved if and

only if f(x) = g(x) for x equal to 2A, 2B, and 2C ; because A + B + C = π, thisis possible only if one of the angles is π

2 while the other two are π4 .


Let Ln denote the nth Lucas number defined by L0 = 2, L1 = 1 and Ln+2 =Ln+1 + Ln for all n ≥ 0. Prove that

L4n + L4

n+1

LnLn+1+L4n+1 + L4

n+3

Ln+1Ln+3+L4n+3 + L4

n

Ln+3Ln≥ 2

3L2n+4.

We received ten correct and complete solutions. We present the solutions of ArkadyAlt, who like most submitters used standard inequalities for a simple proof, and aslightly modified version of the solution by Dionne Bailey, Elsie Campbell, andCharles Diminnie, who made heavier use of the given recursion to find a strongerbound.


Since a4 + b4 ≥ ab(a2 + b2) (as this can be rewritten as (a2 + ab+ b2)(a− b)2 ≥ 0)

and a2 + b2 + c2 ≥ (a+b+c)2

3 for all a, b, c ∈ R, we have

L4n + L4

n+1

LnLn+1+L4n+1 + L4

n+3

Ln+1Ln+3+L4n+3 + L4

n

Ln+3Ln

≥LnLn+1(L2

n + L2n+1)

LnLn+1+Ln+1Ln+3(L2

n+1 + L2n+3)

Ln+1Ln+3+Ln+3Ln(L2

n+3 + L2n)

Ln+3Ln

= 2(L2n + L2

n+1 + L2n+3)

≥ 2(Ln + Ln+1 + Ln+3)2

3=

2(Ln+2 + Ln+3)2

3

=2L2

n+4

3

Solution 2, by Dionne Bailey, Elsie Campbell, and Charles Diminnie.

More generally, we will show that for all n ≥ 0,

L4n + L4

n+1

LnLn+1+L4n+1 + L4

n+3

Ln+1Ln+3+L4n+3 + L4

n

Ln+3Ln> 2L2

n+4.


320/ SOLUTIONS

We can check by hand that this holds for n ≤ 2.

For n ≥ 3 we first use the Arithmetic Mean - Geometric Mean inequality to obtain

x4 + y4 = 2x2y2 + (x2 − y2)2

= 2x2y2 + (x+ y)2(x− y)2

≥ 2x2y2 + 4xy(x− y)2

= xy(2xy + 4(x− y)2)

and hencex4 + y4

xy≥ 2xy + 4(x− y)2.

Using this property and the recursion for the Lucas numbers (multiple times, whennecessary), we get

L4n + L4

n+1

LnLn+1≥ 2LnLn+1 + 4(Ln+1 − Ln)2 = 4L2

n+1 − 6Ln+1Ln + 4L2n,

L4n+1 + L4

n+3

Ln+1Ln+3≥ 2Ln+1Ln+3 + 4(Ln+3 − Ln+1)2 = 8L2

n+1 + 10Ln+1Ln + 4L2n,

L4n+3 + L4

n

Ln+3Ln≥ 2Ln+3Ln + 4(Ln+3 − Ln)2 = 16L2

n+1 + 4Ln+1Ln + 2L2n,

and

2L2n+4 = 18L2

n+1 + 24Ln+1Ln + 8L2n.

Combining these, we obtain

L4n + L4

n+1

LnLn+1+L4n+1 + L4

n+3

Ln+1Ln+3+L4n+3 + L4

n

Ln+3Ln≥ 28L2

n+1 + 8Ln+1Ln + 10L2n

= 2L2n+4 + 10L2

n+1 − 16Ln+1Ln + 2L2n

= 2L2n+4 − 6Ln+1Ln + 10Ln+1Ln−1 + 2L2

n

= 2L2n+4 + 4Ln+1Ln−1 − 6Ln+1Ln−2 + 2L2

n

= 2L2n+4 + 4Ln+1Ln−1 − 4Ln+1Ln−2 + 2LnLn−1 − 2Ln−1Ln−2

= 2L2n+4 + 4Ln+1Ln−3 + L2

n−1

> 2L2n+4.

4063. Proposed by Marcel Chirita.

Let a, b, c be real numbers greater than or equal to 3. Show that

min

Åa2b2 + 3b2

b2 + 27,b2c2 + 3c2

c2 + 27,a2c2 + 3a2

a2 + 27

ã≤ abc

9.


SOLUTIONS /321

We received six submissions all of which were correct. We present a composite ofthe similar solutions by Arkady Alt and Leonard Guigiuc.

Suppose to the contrary that

min

Åa2b2 + 3b2

b2 + 27,b2c2 + 3c2

c2 + 27,a2c2 + 3a2

a2 + 27

ã>abc

9.

Then we have∏cyc

a2b2 + 3b2

b2 + 27>a3b3c3

93, so

∏cyc

a2 + 3

a2 + 27>abc

93.

But since a9 −

a2+3a2+27 = a3−9a2+27a−27

9(a2+27) = (a−3)29(a2+27) ≥ 0, we have a2+3

a2+27 ≤a9 .

Similarly, b2+3b2+27 ≤

b9 and c2+3

c2+27 ≤c9 .

Hence,∏cyc

a2 + 3

a2 + 27>abc

93is a contradiction.


In the plane of a triangle ABC, let Γ be a circle whose centre O is not on thesidelines AB,BC,CA. Let A′, B′, C ′ be the poles of the lines BC,CA,AB withrespect to Γ, respectively. Prove that

OA′ ·B′C ′

OA ·BC=OB′ · C ′A′

OB · CA=OC ′ ·A′B′

OC ·AB.

We received five solutions, all correct, and present the solution by Joel Schlosberg,slightly modified by the editor.

One way to define the pole A′ of the line BC with respect to the circle Γ is byreciprocation, namely A′ is the inverse in Γ of the foot of the perpendicular fromO to BC. [See, for example H.S.M. Coxeter and S.L. Greitzer, Geometry Revisited(The Mathematical Association of America, 1967), Section 6.1.] Conversely, if Dis the inverse of A in Γ, then the polar of A, namely B′C ′, is the line through Dthat is perpendicular to OA. We shall use three immediate consequences of thisdefinition. If r is the radius of Γ, then OA ·OD = r2, or

OA =r2

OD. (1)

Since OB′ ⊥ CA and OC ′ ⊥ AB, ∠B′OC ′ is equal to or supplementary to ∠BAC.Let R be the circumradius of 4ABC. By the law of sines,

BC = 2R sin∠BAC = 2R sin∠B′OC ′. (2)

Finally, since each is the area of 4OB′C ′,

1

2B′C ′ ·OD =

1

2OB′ ·OC ′ sin∠B′OC ′. (3)


322/ SOLUTIONS

Using in turn (1) and (2), then (3), we get

OA′ ·B′C ′

OA ·BC=

OA′ ·B′C ′

(r2/OD) · 2R sin∠B′OC ′=

OA′

2Rr2· B′C ′ ·OD

sin∠B′OC ′=OA′ ·OB′ ·OC ′

2Rr2.

The same reasoning shows thatOB′ · C ′A′

OB · CAand

OC ′ ·A′B′

OC ·ABare also equal to

OA′ ·OB′ ·OC ′

2Rr2.

4065. Proposed by Martin Lukarevski.

Let ABC be a triangle with a, b, c as lengths of its sides and let R, r, s denote thecircumradius, inradius and semiperimeter, respectively. Prove that

1

(s− a)2+

1

(s− b)2+

1

(s− c)2≥ 2

r

Å1

r− 1

R

ã.

We received ten correct and complete solutions. We present the solution by theproposer.

We use the Garfunkel-Bankoff inequality (Problem 825, proposed by J. Garfunkel,solution by L. Bankoff, Crux 9 (1983), p.79 and 10 (1984), p.168) :

tan2 A

2+ tan2 B

2+ tan2 C

2≥ 2− 8 sin

A

2sin

B

2sin

C

2(1)

which by the well-known identity

sinA

2sin

B

2sin

C

2=

r

4R(2)

is equivalent to

tan2 A

2+ tan2 A

2+ tan2 A

2≥ 2− 2r

R.

By another well-known identity, which states that

1

s− a=

1

rtan

A

2, (3)

we have that

1

(s− a)2+

1

(s− b)2+

1

(s− c)2=

1

r2

Åtan2 A

2+ tan2 B

2+ tan2 C

2

ã≥ 1

r2

Å2− 2r

R

ã=

2

r

Å1

r− 1

R

ã,

with equality, as in (1), only for the equilateral triangle.


SOLUTIONS /323


Prove that for a, b, c > 0 and ab+ ac+ bc = 2016,Åa+

1

b

ã2+

Åb+

1

c

ã2+

Åc+

1

a

ã2≥ 20192

2016.

We received 19 solutions all of which are correct. We present a composite of nearlyidentical solutions by Andrea Fanchini and Titu Zvonaru.

We prove the more general result that if a, b, c > 0 such that ab+ bc+ca = k, thenÅa+

1

b

ã2+

Åb+

1

c

ã2+

Åc+

1

a

ã2≥ (k + 3)2

k.

Note first that the trivial inequality x2 + y2 + z2 ≥ xy + yz + zx implies

1

a2+

1

b2+

1

c2≥ 1

ab+

1

bc+

1

ca. (1)

Using (1) together with AM-GM and AM-HM inequalities we then have

(a+1

b)2 + (b+

1

c)2 + (c+

1

a)2 = a2 + b2 + c2 + 2(

a

b+b

c+c

a) +

1

a2+

1

b2+

1

c2

≥ ab+ bc+ ca+ 2 · 3 +1

ab+

1

bc+

1

ca

≥ k + 6 +9

ab+ bc+ ca

=(k + 3)2

k

Editor’s comments. Most of the other solutions used AM-GM, AM-QM and/orCauchy Schwarz Inequalities. It is trivial to see that equality holds if and only if

a = b = c =√3k3 .


Consider a graph G such that between any three vertices in G there are either 0or 2 edges. Classify all such graphs G.

We received seven correct and complete solutions. We present the solution by JoelSchlosberg.

We claim that a graph G satisfies the condition if and only if it is either an edgelessgraph or a complete bipartite graph.

If G is edgeless, any three vertices have zero edges between them, so G triviallysatisfies the condition. If G is a complete bipartite graph, the vertices of G can


324/ SOLUTIONS

be partitioned into two sets S1, S2, such that two vertices are adjacent if and onlyif they are in different sets. If three vertices are in the same set, they have zeroedges between them ; otherwise, two of them are in one set and one is in the other,leading to two edges between them. Thus G satisfies the condition.

Conversely, suppose that G satisfies the condition. If G is not edgeless, then thereexist two vertices v1, v2 with an edge between them. For k = 1, 2, let Sk be the setof vertices that share an edge with vk. Clearly v1 is in S2 but not S1 and v2 in S1

but not S2. If v is a vertex of G different from v1 and v2, then the three verticesv, v1, v2 must have exactly two edges between them, since they cannot have zero.Thus v is in exactly one of S1 or S2. Therefore S1 and S2 form a partition of thevertices of G. Suppose v, w are vertices in the same set, say S1. Then there is noedge between v and w, as otherwise we would have three edges between v, w andv1. Now suppose v ∈ S1 and w ∈ S2. If v = v2 then there is an edge between vand w by the definition of S2. Otherwise consider the three vertices v, w and v2.There is an edge between w and v2 by the definition of S2 and no edge betweenv and v2, as just shown. Therefore there must be an edge between v and w. Thuswe have proven that G is a complete bipartite graph.


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

a+ 2b

2a+ 3b+ c+

b+ 2c

a+ 2b+ 3c+

c+ 2a

3a+ b+ 2c≤ 3

2.

Editor’s comments. We received 25 submissions all of which are correct. However,it was pointed out by Michael Bataille, and Dionne Bailey, Elsie Campbell, andCharles Diminnie that this problem is the same as Crux problem #4016 (by thesame proposer) which appeared on p. 74 of Crux 41 (2). The only difference beingthat in #4016, it was asked to find the maximum value of the given expressionwhile in #4068, it becomes a proof question with the maximum value given. So,it can not be viewed as a “variation”. Two different solutions to #4016 given byArkady Alt and Sefket Arslanagic have appeared on pp. 85-86 of Crux 42 (2). It isinteresting to note that both of them are also among the 25 solvers to the currentproblem but neither made any reference to #4016 !


Let (un)n≥0 be an arithmetic progression with a positive common difference d andwith u1 > 0. Let (xn)n≥0 be a sequence with x0 = 0, x1 = x2 = 1 and

n∑k=1

ukxk = unxn+2 + d(x4 − xn+3)− x2u1, ∀n ≥ 0.

Prove that (xn)n≥0 is the Fibonacci sequence.

There were twelve correct solutions. All were variants of the ones below, with threeusing the closed form of the Fibonacci sums in Solution 2.


SOLUTIONS /325

Solution 1.

When n = 0, the condition is that

0 = u0x2 + d(x4 − x3)− u1x2 = d(x4 − x3 − x2),

whence x4 = x3 + x2. When n = 1, we have that u1x1 = u1x3 − u1x2, whencex3 = x1 + x2. Since x0 = 0 = F0 and x1 = x2 = 1 = F1 = F2, then x3 = F3 andx4 = F4.

For n ≥ 1, we have that

un+1xn+1 = [un+1xn+3 + d(x4 − xn+4)− u1x2]− [unxn+2 + d(x4 − xn+3)− u1x2]

= −dxn+4 + (un+1 + d)xn+3 − (un+1 − d)xn+2,

so thatdxn+4 = d(xn+3 + xn+2) + un+1(xn+3 − xn+2 − xn+1).

We establish the result by induction. Suppose that xk = Fk for 0 ≤ k ≤ n + 3.This is true for n = 1. The foregoing equation establishes that if xk = Fk fork = n+ 1, n+ 2, n+ 3, then dxn+4 = dFn+4 + un+1(0) and xn+4 = Fn+4.

Solution 2.

The following Fibonacci relationships are easily established by induction for n ≥ 1 :

F1 + F2,+ · · ·+ Fn = Fn+2 − 1 andn∑k=1

(k − 1)Fk = (n− 1)Fn+2 − Fn+3 + 3.

As in Solution 1, we show that xk = Fk for 0 ≤ k ≤ 4. Suppose, as an inductionhypothesis, this holds for 1 ≤ k ≤ n + 2. By the foregoing relationships, we havethat

n∑k=1

ukxk =n∑k=1

[u1 + (k − 1)d]Fk

= u1

n∑k=1

Fk + dn∑k=1

(k − 1)Fk

= u1[Fn+2 − 1] + d(n− 1)Fn+2 − dFn+3 + 3d

= Fn+2[u1 + (n− 1)d] + d(3− Fn+3)− u1= unFn+2 + d(F4 − Fn+3)− u1F2.

However, the given condition provides that

n∑k=1

ukxk = unFn+2 + d(F4 − xn+3)− u1F2.

Therefore xn+3 = Fn+3, and the result holds.


326/ SOLUTIONS


Compute

limn→∞

ï1

lnn

Åarctan 1

n+

arctan 2

n− 1+ · · ·+ arctan (n− 1)

2+ arctann

ãò.

We received six correct and complete solutions. We present the solution by JoelSchlosberg.

Let Hn = 1 + 12 + 1

3 + · · ·+ 1n . Since arctanx is an increasing function,

1

lnn

ïarctan 1

n+

arctan 2

n− 1+ · · ·+ arctan(n− 1)

2+ arctann

ò≤ 1

lnn

Åarctann

n+

arctann

n− 1+ · · ·+ arctann

2+ arctann

ã=Hn

lnn· arctann;

and for any positive integer m, if n ≥ m,

1

lnn

ïarctan 1

n+

arctan 2

n− 1+ · · ·+ arctan(n− 1)

2+ arctann

ò≥ 1

lnn

Åarctanm

n−m+ 1+ · · ·+ arctan(n− 1)

2+ arctann

ã≥ 1

lnn

Åarctanm

n−m+ 1+ · · ·+ arctanm

2+ arctanm

ã=Hn−m+1

lnn· arctanm.

It is well known that Hn is asymptotic to lnn and limx→∞ arctanx = π/2. The-refore,

limn→∞

Hn

lnn· arctann =

π

2

and

limm→∞

Ålimn→∞

Hn−m+1

lnn· arctanm

ã= lim

m→∞

Ålimn→∞

ln(n−m+ 1)

lnn· arctanm

ã= limm→∞

arctanm =π

2

so by the squeeze theorem,

limn→∞

1

lnn

Åarctan 1

n+ · · ·+ arctan(n− 1)

2+ arctann

ã=π

2.


358/ SOLUTIONS

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



Prove that if a, b, c ∈ (0, 1), then aa+1bb+1cc+1 < e2(a+b+c)−6.

There were 14 correct solutions and one incorrect submission. We present the so-lution submitted by various solvers.

Letf(x) = 2(x− 1)− (x+ 1) lnx = 2(x− 1)− x lnx− lnx

for 0 < x < 1. The derivative

f ′(x) = 1− x−1 − lnx = ln(x−1)− (x−1 − 1)

is negative and f(1) = 0, so that f(x) > 0 on (0, 1). Therefore (x+1) lnx < 2(x−1)for 0 < x < 1.

Thus(a+ 1) ln a+ (b+ 1) ln b+ (c+ 1) ln c < 2(a+ b+ c)− 6.

Exponentiating yields the result.

Editor’s Comments. Several solvers used the function in the solution ; anotherpopular function studied was lnx− 2(x− 1)(x+ 1)−1 = lnx+ 4(x+ 1)−1 − 2 or aclose relative. Three solvers noted that the function (x + 1) lnx was concave andapplied Jensen’s inequality. One respondent took a stroll down the garden pathwith the following argument.

Let L = aa+1bb+1cc+1. Since lnx < x− 1 for 0 < x < 1,

lnL = (a+ 1) ln a+ (b+ 1) ln b+ (c+ 1) ln c < 2(ln a+ ln b+ ln c)

< 2((a− 1) + (b− 1) + (c− 1)) = 2(a+ b+ c)− 6.

Exponentiating yields the desired result.


Let a, b be distinct positive real numbers and A = a+b2 , G =

√ab, L = a−b

ln a−ln b .Prove that

L

G>

4A+ 5G

A+ 8G.

There were five correct solutions, of which we present two.

Crux Mathematicorum, Vol. 42(8), October 2016

SOLUTIONS /359


This inequality is a refinement of the known inequality L > G. Wolog, let a > band set x =

√a, y =

√b. The inequality can be rewritten as

lnx− ln y

x2 − y2<

1

4xy· x

2 + y2 + 16xy

2x2 + 2y2 + 5xy.

Setting t = x/y converts it to

4 ln t

t2 − 1<

1

t· t

2 + 16t+ 1

2t2 + 5t+ 2.

Since1

t· t

2 + 16t+ 1

2t2 + 5t+ 2=

1

2

Å1

t+

27

2t2 + 5t+ 2

ã,

it all boils down to proving for t > 1 the inequality

8 ln t

t2 − 1<

1

t+

27

2t2 + 5t+ 2.

Recall Simpson’s 38 Rule for numerical integration of a C4-function f on a closed

interval [u, v] :∫ v

u

f(s)ds =v − u

8

Åf(u) + 3f

Å2u+ v

3

ã+ 3f

Åu+ 2v

3

ã+ f(v)

ã− (v − u)5

6480f (4)(ξ)

for some ξ ∈ (u, v). Taking f(s) = 1/s, u = 1 and v = t > 1, we obtain

ln t =t− 1

8

Å1 +

9

2 + t+

9

1 + 2t+

1

t

ã− 24(t− 1)5

6480ξ2

and so

ln t <t− 1

8

Å1 +

1

t+

27(t+ 1)

2t2 + 5t+ 2

ã=t2 − 1

8

Å1

t+

27

2t2 + 5t+ 2

ã.

This leads to the desired inequality.


Assume a > b and let t =√a/b. Then as in the previous solution, we have to

establish that

4 ln t <(t2 − 1)(t2 + 16t+ 1)

t(t+ 2)(2t+ 1).

Let

h(t) =(t2 − 1)(t2 + 16t+ 1)

t(t+ 2)(2t+ 1)− 4 ln t

=t

2− 27

2(t+ 2)− 27

4(2t+ 1)− 1

2t+

27

4− 4 ln t.


360/ SOLUTIONS

Then

h′(t) =1

2+

27

2(t+ 2)2+

27

2(2t+ 1)2+

1

2t2− 4

t

=2(t2 + t+ 1)(t− 1)4

t2(t+ 2)2(2t+ 1)2> 0

for t > 1. Since h(t) > h(1) = 0 for t > 1, the inequality follows.

4073. Proposed by Daniel Sitaru.

Solve the following system :®sin 2x+ cos 3y = −1,√

sin2 x+ sin2 y +√

cos2 x+ cos2 y = 1 + sin(x+ y).

The solution from Michel Bataille was the only one of the 2 submissions that wascomplete and correct. We present his solution.

We first show that the second equation is equivalent to x+ y ≡ π2 (mod 2π).

If x + y ≡ π2 (mod 2π), then sin2 y = cos2 x and cos2 y = sin2 x. It immediately

follows that both sides of the equation equal 2. Conversely, if the equation holds,then squaring gives

2»

(sinx cosx− sin y cos y)2 + sin2(x+ y) = 2 sin(x+ y)− (1− sin2(x+ y)),

and therefore

2 sin(x+ y) ≤ 2»

sin2(x+ y) ≤ 2»

(sinx cosx− sin y cos y)2 + sin2(x+ y)

= 2 sin(x+ y)− (1− sin2(x+ y)) ≤ 2 sin(x+ y).

Thus, equality must hold throughout and in particular sin(x+y) ≥ 0 and sin2(x+y) = 1. We deduce that x+ y ≡ π

2 (mod 2π).

Since cos 3(π2 − x

)= − sin 3x, we are led to seek the solutions to the equation

f(x) = 1 where f(x) = sin 3x− sin 2x. Note that f(−π2)

= 1 so that the numbers−π2 + 2kπ (k ∈ Z) are solutions. For other solutions note that f is odd and2π-periodic ; consequently, we may restrict the study of f to the interval [0, π]and look for x satisfying either f(x) = 1 or f(x) = −1 (the latter since thenf(−x) = 1). Consider first the interval [0, π2 ). We have f(0) = 0 and if x ∈ (0, π2 ),then sin 2x > 0 and so f(x) < 1.

• x ∈ (0, π3 ] : sin 3x > 0 for x between 0 and π3 , hence f(x) > −1 ; since

f(π3

)> −1, there is no x ∈ (0, π3 ] such that f(x) = −1.

• x ∈ (π3 ,π2 ) : f ′′(x) = 4 sin 2x−9 sin 3x > 0, hence f ′(x) = 3 cos 3x−2 cos 2x is

nondecreasing on the interval (π3 ,π2 ). For some x1 ∈ (π3 ,

π2 ), we have f ′(x) ≤ 0


SOLUTIONS /361

for x ∈ (π3 , x1] and f ′(x) > 0 for x ∈ (x1,π2 ). Since f

(π3

)= −

√3

2 andf(π2

)= −1, we have f(x1) < −1 and f(α) = −1 for a unique α in (π3 ,

π2 ).

In a similar way we treat the interval (π2 , π]. We have f(π) = 0 and if x ∈ (π2 , π),then sin 2x < 0, hence f(x) > −1.

• x ∈ (π2 ,3π4 ) : f ′(x) > 0 and so f is increasing from −1 to 1 +

√2

2 . Thus,f(β) = 1 for a unique β ∈ (π2 ,

3π4 ).

• x ∈ ( 5π6 , π) : f is decreasing from 1 +

√3

2 to 0, hence f(γ) = 1 for a uniqueγ of ( 5π

6 , π).

• x ∈ [ 3π4 ,

5π6 ] : Resorting to f ′′(x), we see that f ′(x) decreases from positive

to negative so that f(x) > 1.

In conclusion, on the interval [−π, π] the solutions (x, y) of the system are thepairs (

−π2, π),(−α, π

2+ α

),(β,

π

2− β

), and

(γ,

π

2− γ).

All other solutions are obtained by adding multiples of 2π to x or y.


Consider the triangle ABC with the following measures :

Show that a+ b = c ; that is, |AE|+ |AC| = |AB|.

We received 16 solutions, all correct. We feature the solution of C.R. Pranesacharthat is typical of the approach used by most solvers.

Because the angles of ∆BCE sum to 180◦, we see that ∠BED = 100◦ and itssupplement ∠BEA = 80◦. It follows that in ∆BEA we also have ∠BAE = 80◦,so that BE = AB = c and

AE/2

AB= sin

20◦

2,

or

a = 2c sin 10◦.


362/ SOLUTIONS

Further, by the Sine Rule applied to triangle BCE,

CE = BEsin 10◦

sin 30◦= 2c sin 10◦ = a.

This means that ∆AEC is also isosceles (with EA = EC = a), and because itsexterior angle at E equals 40◦, its interior angles at A and C must each be 20◦ ;thus

b/2

a= cos 20◦ or b = 2(2c sin 10◦) · cos 20◦.

The given relation a+ b = c now translates to

2 sin 10◦ + 4 sin 10◦ cos 20◦ = 1.

This is easy to prove :

lhs = 2 sin 10◦ + 2(sin 30◦ − sin 10◦) = 1 = rhs.


Prove that in any triangle ABC with BC = a, CA = b, AB = c the followinginequality holds :

3√abc ·

√a2 + b2 + c2 ≥ 4[ABC],

where [ABC] is the area of triangle ABC.

We received 16 correct solutions and we present the solution by Martin Lukarevski.

The inequality can be sharpened to

3√abc ·

√ab+ bc+ ca ≥ 4[ABC].

We use the inequality3√abc ≥

»4[ABC]/

√3,

which is equivalent to the well-known inequality

sinA sinB sinC ≤ 3√

3

8,

and Gordon’s inequality (4.5 in O. Bottema, R. Z. Djordjevic, R. R. Janic, D.S. Mitrinovic, P. M. Vasic, Geometric inequalities, Groningen, Wolters-Noordhoff,1969)

√ab+ bc+ ca ≥

»4[ABC]

√3.

Hence3√abc ·

√a2 + b2 + c2 ≥ 3

√abc ·

√ab+ bc+ ca ≥ 4[ABC].

Editor’s Comments. Many of the solutions were rather similar in nature as mostverifications were the result of combining existing inequalities. In fact, MartinLukarevski submitted two solutions of which his second is presented above.


SOLUTIONS /363


Prove that (x2 + y2 + z2)3 ≥ (x3 + y3 + z3 + 3(√

3 − 1)xyz)2 for all nonnegativereals x, y, and z.

We received seven submissions, six of which were correct. We present the solutionby Michel Bataille, expanded slightly by the editor.

Note first that equality holds if x = y = z = 0. Now suppose x+ y + z > 0. Thenby homogeneity we may assume that x + y + z = 1. Let m = xy + yz + zx andk = xyz. Then x2 + y2 + z2 = 1− 2m and

x3 + y3 + z3 = 3xyz + (x+ y + z)(x2 + y2 + z2 − xy − yz − zx) = 1− 3m+ 3k.

The given inequality is equivalent, in succession, to

(1− 2m)3 ≥ (1− 3m+ 3√

3k)2

1− 6m+ 12m2 − 8m3 ≥ 1 + 9m2 + 27k2 − 6m+ 6√

3k − 18√

3mk

3m2 + 18√

3mk − 8m3 − 27k2 − 6√

3k ≥ 0. (1)

Note that 1− 3m ≥ 0 since x2 + y2 + z2 ≥ xy + yz + zx. We set u =√

1− 3m som = 1

3 (1− u2) and then (1) becomes

8

27(1− u2)3 + 27k2 + 6

√3k ≤ 1

3(1− u2)2 + 6

√3(1− u2)k

8

27(1− 3u2 + 3u4 − u6) + 27k2 + 6

√3k ≤ 1

3(1− 2u2 + u4) + 6

√3k − 6

√3ku2

27 · 6√

3ku2 + (27k)2 ≤ 1 + 6u2 − 15u4 + 8u6. (2)

We now apply the following known result :

27k ≤ (1− u)2(1 + 2u) = 1− 3u2 + 2u3. (3)

(See the article “On a class of three-variable Inequalities” by Vo Quoc Ba Can,Mathematical Reflections, 2007, issue 2, and the proof by Paolo Perfetti in hissolution to Crux Problem 3663, 38 (7), pp. 291-292.)

Using (3), we see that in order to establish (2), it suffices to prove the inequality :

6√

3u2(1− u)2(1 + 2u) + (1− u)4(1 + 2u)2 ≤ 1 + 6u2 − 15u4 + 8u6. (4)

After straightforward computations, we see that (4) is successively equivalent to :

(1− u)2(6√

3u2(1 + 2u) + (1− u)2(1 + 2u)2) ≤ (1− u)2(1 + 2u+ 9u2 + 16u3 + 8u4)

1 + 2u+ 9u2 + 16u3 + 8u4 − (6√

3u2 + 12√

3u3 + 1 + 2u− 3u2 − 4u3 + 4u4) ≥ 0

((12− 6√

3)u2 + (20− 12√

3)u3 + 4u4) ≥ 0

2u2 + (10− 6√

3)u+ 6− 3√

3 ≥ 0,

which is true since (10− 6√

3)2− 8(6− 3√

3) = 32(5− 3√

3) < 0. Hence (4) is trueand our proof is complete.


364/ SOLUTIONS


Let ABC be a triangle. Prove that

sinA

2· sinB · sinC + sinA · sin B

2· sinC + sinA · sinB · sin C

2≤ 9

8.

We received 15 submissions, of which 14 were correct and complete. We presentthe solution by Phil McCartney.

Let a = π−A2 , b = π−B

2 , c = π−C2 . Then a+ b+ c = π and a, b and c are in

(0, π2

).

We have ∑cyc

sinA

2sinB sinC =

∑cyc

cos a sin(π − 2b) sin(π − 2c)

=∑cyc

cos a sin(2b) sin(2c)

= 4 cos a cos b cos c ·∑cyc

sin a sin b

Hence it suffices to show that

cos a cos b cos c ≤ 1

8and

∑cyc

sin a sin b ≤ 9

4. (†)

Since cos(t) is a concave function on(0, π2

), the AM-GM inequality followed by

Jensen’s inequality yields :

cos a cos b cos c ≤Å

cos a+ cos b+ cos c

3

ã3

≤ cos3

Åa+ b+ c

3

ã=

1

8,

proving the first of the two inequalities in (†).

By Cauchy’s inequality,∑cyc

sin a sin b ≤∑cyc

sin2 a ; so, in order to conclude the

second inequality also holds, it suffices to prove that∑cyc

sin2 a ≤ 94 . However, one

can show that∑cyc

sin2 a = 2(1 + cos a cos b cos c) : using the fact that a+ b+ c = π,

and hence sin(a) = sin(b+ c), we have

sin2 a = (sin b cos c+ cos b sin c)2

= sin2 b cos2 c+ 2 sin b cos b cos c sin c+ cos2 b sin2 c

= (1− cos2 b) cos2 c+ 2 sin b cos b cos c sin c+ cos2 b(1− cos2 c)

= cos2 b+ cos2 c− 2 cos2 b cos2 c+ 2 cos b cos c sin b sin c

= cos2 b+ cos2 c− 2 cos b cos c cos(b+ c)

= 1− sin2 b+ 1− sin2 c+ 2 cos b cos c cos a,

which can be rearranged to sin2 a+sin2 b+sin2 c = 2+2 cos a cos b cos c as claimed.


SOLUTIONS /365

Finally, using the first inequality in (†), we can conclude that∑cyc

sin2 a ≤ 2+ 14 = 9

8 ,

showing the second inequality in (†).

Editor’s Comments. The inequalities in (†) are known and can be found in O. Bot-tema et al., Geometric inequalities, Groningen, Wolters-Noordhoff, 1969.


Given θ such that π3 ≤ θ ≤ 5π

3 , let M0 be a point of a circle with centre O andradius R and Mk its image under the counterclockwise rotation with centre O andangle kθ. If M is the point diametrically opposite to M0 and n is a positive integer,show that

n∑k=0

MMk ≥ (2n+ 1) · R2.

We received two submissions, both correct, and feature the solution by AN-anduudProblem Solving Group.

We can assume that R = 1,M0 = e0 = 1, and M = −1 ; then Mk = eikθ,k = 1, 2, . . . , n. Let eiθ = z, so that Mk = zk, and

MMk = | − 1− eikθ| = |1 + zk| ≤ 1 + |z|k = 2.

Thus we haven∑k=0

MMk =n∑k=0

|1 + zk| = 1

2

n∑k=0

2 · |1 + zk|

≥ 1

2

n∑k=0

|1 + zk|2 =1

2

n∑k=0

(1 + zk)(1 + zk)

=1

2

n∑k=0

(1 + zk)(1 + z−k) =1

2

n∑k=0

(2 + (zk + z−k))

=n∑k=0

Ç1 +

zk + z−k

2

å=

n∑k=0

(1 + cos kθ)

= n+1

2+

(1 +

1

2+

n∑k=1

cos kθ

)

=2n+ 1

2+

Ç1 +

sin(n+ 1

2

)θ

2 sin θ2

å=

2n+ 1

2+

2 sin θ2 + sin

(n+ 1

2

)θ

2 sin θ2Å

π

6≤ θ

2≤ 5π

6⇒ sin

θ

2≥ 1

2⇒ 2 sin

θ

2≥ 1

ã≥ 2n+ 1

2+

1 + sin(n+ 1

2

)θ

2 sin θ2

≥ 2n+ 1

2.


366/ SOLUTIONS


Let x, y, z > 0 and x+ y + z = 2016. Prove that :

x

…yz

y + 2015z+ y

…xz

z + 2015x+ z

…xy

x+ 2015y≤ 2016√

3.

We received eleven solutions. We present 2 solutions.


We prove the general case. Let x+ y + z = t, where t ≥ 1. We have

yz

y + (t− 1)z≤ (t− 1)y + z

t2. (1)

Indeed, the inequality (1) is equivalent to

(t− 1)y2 + (t− 1)z2 + ((t− 1)2 + 1)yz ≥ t2yz ⇐⇒ (t− 1)(y − z)2 ≥ 0.

Using (1), the Cauchy-Schwarz Inequality and the known inequality

(x+ y + z)2 ≥ 3(xy + yz + zx),

we obtain

x…

yz

y + (t− 1)z+ y…

xz

z + (t− 1)x+ z…

xy

x+ (t− 1)z

≤ x

t

√(t− 1)y + z +

y

t

√(t− 1)z + x+

z

t

√(t− 1)x+ y

=1

t

Ä√x√

(t− 1)xy + zx+√y√

(t− 1)yz + xy +√z√

(t− 1)zx+ yzä

≤ 1

t

√(x+ y + z)[(t− 1)xy + zx+ (t− 1)yz + xy + (t− 1)zx+ yz]

=1

t

√t2(xy + yz + zx)

≤ 1

t· t…

(x+ y + z)2

3=

t√3

The equality holds if and only if x = y = z = t/3.

Solution 2, by Dionne Bailey, Elsie Campbell, and Charles Diminnie.

We will prove the following slight generalization of the given problem :

If x, y, z > 0 and x+ y + z = S + 1, then

x

…yz

y + Sz+ y

…zx

z + Sx+ z

…xy

x+ Sy≤ S + 1√

3.


SOLUTIONS /367

To begin, we invoke the general form of the Arithmetic - Geometric Mean Inequa-lity which states that if a, b, α, β > 0 and α+ β = 1, then

aα · bβ ≤ αa+ βb, (2)

with equality if and only if a = b. It follows from (2) that

(y + Sz) (Sy + z) = (S + 1)2Å

1

S + 1y +

S

S + 1z

ãÅS

S + 1y +

1

S + 1z

ã≥ (S + 1)

2Äy

1S+1 z

SS+1

ä Äy

SS+1 z

1S+1

ä= (S + 1)

2yz,

and hence,

x

…yz

y + Sz≤ x

S + 1

√Sy + z. (3)

Further, equality is attained in (3) if and only if y = z.

Similar arguments show that

y

…zx

z + Sx≤ y

S + 1

√Sz + x, (4)

with equality if and only if z = x, and

z

…xy

x+ Sy≤ z

S + 1

√Sx+ y, (5)

with equality if and only if x = y.

Since f (t) =√t is strictly concave on (0,∞), we utilize conditions (3), (4), (5),

the constraint equation x+ y + z = S + 1, and Jensen’s Inequality to obtain

x

…yz

y + Sz+ y

…zx

z + Sx+ z

…xy

x+ Sy

≤ x

S + 1

√Sy + z +

y

S + 1

√Sz + x+

z

S + 1

√Sx+ y

≤

x (Sy + z) + y (Sz + x) + z (Sx+ y)

S + 1

=√xy + yz + zx

≤

(x+ y + z)

2

3

=S + 1√

3,

with equality if and only if x = y = z =S + 1

3.


368/ SOLUTIONS

4080. Proposed by Alina Sıntamarian and Ovidiu Furdui.

Let a, b ∈ R, with ab > 0. Calculate∫ ∞0

x2e−(ax− bx )

2

dx.

We received nine submissions of which five were correct and complete solutions.We present the solution by Michel Bataille.

We show that the value of the given integral I is

I =

√π(1 + 2ab)

4|a|3.

Recall that∫∞

0e−t

2

dt =√π

2 . For later use, we also calculate J =∫∞

0t2e−t

2

dt.For X > 0, integrating by parts, we obtain∫ X

0

t2e−t2

dt =1

2

Çî−te−t

2óX0

+

∫ X

0

e−t2

dt

å=

1

2

Ç−Xe−X

2

+

∫ X

0

e−t2

dt

åand letting X →∞,

J =1

2

∫ ∞0

e−t2

dt =

√π

4.

The equation ax− bx = y has a unique positive solution for x = 1

2a (y+ε√y2 + 4ab),

where ε = 1 if a, b > 0 and ε = −1 if a, b < 0. The change of variables

x =1

2a(y + ε

√y2 + 4ab), dx =

ε

2a· y + ε

√y2 + 4ab√

y2 + 4abdy

yields

I =1

8a3

∫ ∞−∞

(y + ε√y2 + 4ab)3√

y2 + 4abe−y

2

dy.

We expand the non-exponential factor in the integrand as

(y + ε√y2 + 4ab)3√

y2 + 4ab=

y3√y2 + 4ab

+ 3εy2 + 3y√y2 + 4ab+ ε(y2 + 4ab).

Note the behaviour of the first and third terms on the right-hand side :

|y|√y2 + 4abe−y

2

∼ |y|3√y2 + 4ab

e−y2

∼ y2e−y2

as y →∞

It follows that the integrals∫ ∞−∞

y3√y2 + 4ab

e−y2

dy and

∫ ∞−∞

3y√y2 + 4ab e−y

2

dy

exist and are therefore zero, since each integrand is odd. Thus,

I =1

8a3

Å6εJ + 2εJ + 8εab

∫ ∞0

e−y2

dy

ã=

√π(1 + 2ab)

4|a|3.


SOLUTIONS /405




Determine all A,B ∈M2(R) such that:A2 +B2 =

Ç22 44

14 28

å,

AB +BA =

Ç10 20

2 4

å.

We received 17 correct solutions and will feature the solution by Joseph DiMuro.

Summing the two equations, we obtain:

(A+B)2 = A2 +AB +BA+B2 =

Å32 6416 32

ã.

We can diagonalize this matrix in order to find its square roots:

(A+B)2 = PDP−1 =

Å2 21 −1

ãÅ64 00 0

ãÅ1/4 1/21/4 −1/2

ã,

A+B = PD1/2P−1 =

Å2 21 −1

ãÅ±8 00 0

ãÅ1/4 1/21/4 −1/2

ã= ±

Å4 82 4

ã.

We can also subtract the original two equations to obtain:

(A−B)2 = A2 −AB −BA+B2 =

Å12 2412 24

ã.

As before, we diagonalize this matrix:

(A−B)2 = PDP−1 =

Å1 21 −1

ãÅ36 00 0

ãÅ1/3 2/31/3 −1/3

ã,

A−B = PD1/2P−1 =

Å1 21 −1

ãÅ±6 00 0

ãÅ1/3 2/31/3 −1/3

ã= ±Å

2 42 4

ã.


406/ SOLUTIONS

Now we have the two equations

A+B = ±Å

4 82 4

ã, A−B = ±

Å2 42 4

ã,

which can easily be solved to produce four possible pairs of matrices for A and B.One solution is

A =

Å3 62 4

ã, B =

Å1 20 0

ã.

The other solutions may be obtained by interchanging A and B, and/or replacingA and B with their negatives.


Let ABC be a right-angle triangle with ∠A = 90◦ and BC = a, AC = b and AB =c. Consider the Fibonacci sequence Fn with F0 = F1 = 1 and Fn+2 = Fn+1 + Fnfor all non-negative integers n. Prove that

F 2m

(bFn + cFp)2+

F 2n

(bFp + cFm)2+

F 2p

(bFm + cFn)2≥ 3

2a2

for all non-negative integers m, n, p.

We received 8 correct solutions and present the solution by Adnan Ali.

From the Cauchy-Schwarz Inequality,(b2 + c2)(F 2

k + F 2` ) = a2(F 2

k + F 2` ) ≥ (bFk + cF`)

2, for all k, ` ≥ 0. Thus,

F 2m

(bFn + cFp)2+

F 2n

(bFp + cFm)2+

F 2p

(bFm + cFn)2≥

F 2m

a2(F 2n + F 2

p )+

F 2n

a2(F 2p + F 2

m)+

F 2p

a2(Fm + F 2n)≥

3

2a2,

where the last inequality follows from Nesbitt’s Inequality. Equality holds iffFm = Fn = Fp and b = c.

Editor’s Comments. As solvers pointed out, the fact that the Fn’s were Fibonaccinumbers was irrelevant; it was only necessary that they were nonnegative.


Calculate

limn→∞

1

n√n

∫ n

0

x

1 + n cos2 xdx.

We received 10 solutions, of which 6 were correct and complete. We present thesolution by Michel Bataille.

Crux Mathematicorum, Vol. 42(9), November 2016

SOLUTIONS /407

We show that the required limit is1

2.

Let fn(x) = 11+n cos2 x .

The π-periodicity of fn and the change of variables x = tan−1(t), dx = dt1+t2 easily

yield ∫ (2k+1)π/2

(2k−1)π/2fn(x) dx =

∫ π/2

−π/2fn(x) dx =

∫ ∞−∞

dt

n+ 1 + t2=

π√n+ 1

for any k, n ∈ N.

This said, for every n ∈ N with n ≥ 2, let pn =⌊nπ + 1

2

⌋and In =

∫ n0xfn(x) dx.

Then, (2pn − 1)π2 ≤ n < (2pn + 1)π2 and

In =

∫ π/2

0

xfn(x) dx+

pn−1∑k=1

∫ (2k+1)π/2

(2k−1)π/2xfn(x) dx+

∫ n

(2pn−1)π2xfn(x) dx.

Clearly,

0 ≤∫ π/2

0

xfn(x) dx ≤ π

2· 1

2

∫ π/2

−π/2fn(x) dx =

π2

4√n+ 1

and for k ∈ {1, 2, . . . , pn − 1},

(2k − 1)π

2· π√

n+ 1≤∫ (2k+1)π/2

(2k−1)π/2xfn(x) dx ≤ (2k + 1)

π

2· π√

n+ 1.

Similarly,

0 ≤∫ n

(2pn−1)π2xfn(x) dx ≤ n

∫ n

(2pn−1)π2fn(x) dx ≤ nπ√

n+ 1.

Thus,

π2

2√n+ 1

pn−1∑k=1

(2k − 1) ≤ In ≤π2

4√n+ 1

+π2

2√n+ 1

pn−1∑k=1

(2k + 1) +nπ√n+ 1

so that

π2(pn − 1)2

2√n+ 1

≤ In ≤π√n+ 1

(π4

+π

2· p2n + n

)=

πp2n√n+ 1

Åπ

2+

π

4p2n+

n

p2n

ã.

Since pn ∼ nπ as n→∞, we obtain

In ∼π2p2n

2√n+ 1

∼ n√n

2

as n→∞. The result follows.


408/ SOLUTIONS


In the plane, let Γ be a circle and A,B be two points not on Γ. Determine whenMAMB is not independent of M on Γ and, in these cases, construct with ruler andcompass I and S on Γ such that

IA

IB= inf

ßMA

MB: M ∈ Γ

™and

SA

SB= sup

ßMA

MB: M ∈ Γ

™.

We feature the proposer’s solution; we received no others.

Because A is not on Γ, inversion in the circle with centre A and radius AB takesΓ to another circle, call it Γ′. For any point M on Γ, this inversion takes the pairof points M,B to another pair M ′, B, whose distances satisfy

MB =AB2 ·M ′BAM ′ ·AB

=AM ·AM ′ ·M ′B

AM ′ ·AB= MA

M ′B

AB;

consequently,MA

MB=

AB

BM ′. (1)

From (1), MAMB is independent of M on Γ if and only if BM ′ is constant. This

occurs if and only if B is the centre of Γ′; that is, if and only if A and B arean inverse pair with respect to Γ. Otherwise, let the diameter of Γ′ through Bintersect Γ′ at C and D with BC > BD. Then MA

MB is minimal when BM ′ is

maximal; that is, when M ′ = C; MAMB is maximal when BM ′ is minimal, in which

case M ′ = D. Thus, I coincides with C ′ (the image of C under our inversion),and S coincides with D′. The construction of I and S is immediate once Γ′ hasbeen drawn. The circle Γ′ can be readily constructed from the inverses of threepoints of Γ (as in the figure):

Editor’s Comments. For any two fixed points A and B, the locus of points Mfor which MA

MB is constant is called the circle of Apollonius; inversion in that circleinterchanges A and B. See, for example H.S.M. Coxeter and S.L. Greitzer, Geom-etry Revisited (Mathematical Association of America, 1967), exercise 5.4.1, pages114 and 172. Also, Theorem 5.41 there provides the distance formula used aboveto obtain (1).


SOLUTIONS /409

4085. Proposed by Jose Luis Dıaz-Barrero. Correction.

Let ABC be an acute triangle. Prove that

4

»sin(cosA) · cosB + 4

»sin(cosB) · cosC + 4

»sin(cosC) · cosA <

3√

2

2.

We received eight submissions, six of which are correct. We present the solutionby Titu Zvonaru.

It is well known that cosA + cosB + cosC ≤ 32 [Item 2.16 on p.22 of the book

Geometric Inequalities by O. Bottema et al; Groningen, 1969]. Using this, togetherwith the facts that sinx < x for 0 < x < π

2 , xy + yz + zx ≤ x2 + y2 + z2, and(x+ y + z)2 ≤ 3(x2 + y2 + z2) we then have∑

cyc

4

»sin(cosA) · cosB <

∑cyc

4√

cosA · cosB ≤∑cyc

√cosA

≤»

3(cosA+ cosB + cosC) ≤…

3(3

2) =

3√

3

2.

Editor’s comments. Arkady Alt proved the stronger result that the given upper

bound could be replaced by 3 4

»12 sin 1

2 which is less than 3√3

2 since sin 12 < 1

2 .This new upper bound is attained if and only if the triangle is equilateral. Hisproof used the Cauchy-Schwarz Inequality, concavity of the functions

√sinx and√

cosx, Jensen’s Inequality as well as the fact that∑

cosA = 1 + rR and the

Euler’s Inequality 2r ≤ R.


Let be f : [0, 1]→ R; f twice differentiable on [0, 1] and f ′′(x) < 0 for all x ∈ [0, 1].Prove that

25

∫ 1

15

f(x)dx ≥ 16

∫ 1

0

f(x)dx+ 4f(1).

We received seven solutions and present two of them.

Solution 1, by AN-anduud Problem Solving Group.

From the given conditions, f is concave on [0, 1]. Using Hermite-Hadamard’sinequality we get

16

∫ 1

15

f(x)dx+ 9

∫ 1

15

f(x)dx ≥ 16 ·∫ 1

15

f(x)dx+ 9 ·1− 1

5

2·Åf(1) + f

Å1

5

ãã= 16

∫ 1

15

f(x)dx+18

5f(1) +

18

5f

Å1

5

ã.

On the other hand, we have

f

Å1

5

ã= f

Å1

9· 1 +

8

9· 1

10

ã≥ 1

9f(1) +

8

9· fÅ

1

10

ã,


410/ SOLUTIONS

so 18

5f

Å1

5

ã≥ 2

5f(1) +

16

5f

Å1

10

ã.

From here, we get

25

∫ 1

15

f(x)dx ≥ 16

∫ 1

15

f(x)dx+ 4f(1) +16

5f

Å1

10

ã.

Using Hermite-Hadamard’s inequality, we get

f

Å1

10

ã= f

Ç15 + 0

2

å≥ 1

15 − 0

∫ 15

0

f(x)dx ⇐⇒ 1

5f

Å1

10

ã≥∫ 1

5

0

f(x)dx.

Hence, we get

25

∫ 1

15

f(x)dx ≥ 16

∫ 1

0

f(x)dx+ 4f(1).


In

∫ 1

15

f(x) dx, we make the substitution x→ 5x− 1

4and clear fractions to get

25

∫ 1

15

f(x) dx = 20

∫ 1

0

f

Å4x+ 1

5

ãdx.

We need to prove

20

∫ 1

0

f

Å4x+ 1

5

ãdx ≥ 16

∫ 1

0

f(x) dx+ 4f(1) ⇐⇒∫ 1

0

f

Å4x+ 1

5

ãdx ≥ 4

5

∫ 1

0

f(x) dx+1

5f(1) ⇐⇒∫ 1

0

f

Å4x+ 1

5

ãdx ≥

∫ 1

0

ï4

5f(x) +

1

5f(1)

òdx.

But f ′′(x) < 0 ∀x ∈ [0, 1], so f is concave on [0, 1] and from here

f

Å4x+ 1

5

ã≥ 4

5f(x) +

1

5f(1).

Integrating, we conclude that∫ 1

0

f

Å4x+ 1

5

ãdx ≥

∫ 1

0

ï4

5f(x) +

1

5f(1)

òdx.

Editor’s Comments. Henry Ricardo observed that this problem appears as problemMA 110 (with solution) in the Daniel Sitaru’s book Math Phenomenon, publishedin English by the Romanian publisher Editura Paralela 45 in 2016.


SOLUTIONS /411

4087. Proposed by Lorian Saceanu.

If S is the area of triangle ABC, prove that

ma(b+ c) + 2m2a ≥ 4S sinA,

where b and c are the lengths of sides that meet in vertex A, and ma is the lengthof the median from that vertex; furthermore, equality holds if and only if b = cand ∠A = 120◦.

We received seven correct submissions and present the solution by Leonard Giugiuc.

Let A′ be the reflection of A in the midpoint M of BC. Because ∆AMC ∼=∆A′MB, we have

AA′ = 2ma, A′B = b, ∠A′BA = π −A, and [A′AB] = [ABC] = S

(where the square brackets denote area). Let m = 2ma and denote by r′, R′,and s′ (= m+b+c

2 ) the inradius, circumradius, and semiperimeter, respectively, of∆A′AB. We need to prove that

m(b+ c) +m2 ≥ 8S sin∠A′BA,

which is equivalent, in turn, to

m(m+ b+ c) ≥ 8S sin(π −A)

2mS

r′≥ 8S sinA

m

sinA≥ 4r′

R′ ≥ 2r′.

But the final line is Euler’s inequality applied to ∆A′AB, which completes theproof. Equality holds for Euler’s inequality if and only if ∆A′AB is equilateral,which implies that b = c and ∠A = 120◦, as desired.

4088. Proposed by Ardak Mirzakhmedov.

Let a, b and c be positive real numbers such that a2b + b2c + c2a + a2b2c2 = 4.Prove that

a2 + b2 + c2 + abc(a+ b+ c) ≥ 2(ab+ bc+ ca).

We received four submissions all of which are correct. We present the solution bythe proposer, expanded by the editor with some details.

We first show that the given condition implies

a

2a+ bc2+

b

2b+ ca2+

c

2c+ ab2= 1 (1)

or

a(2b+ ca2)(2c+ ab2) + b(2c+ ab2)(2a+ bc2) + c(2a+ bc2)(2b+ ca2)

= (2a+ bc2)(2b+ ca2)(2c+ ab2).(2)


412/ SOLUTIONS

Let S and P denote the left side and the right side of (2), respectively. Then bystraightforward computations, we find

S =∑cyc

a(4bc+ 2ab3 + 2c2a2 + a3b2c)

= 12abc+ 4(a2b3 + b2c3 + c2a3) + abc(a3b+ b3c+ c3a)

= 12abc+ 4(a2b3 + b2c3 + c2a3) + abc(4− a2b2c2)

= 16abc+ 4(a2b3 + b2c3 + c2a3)− a3b3c3

and

P = (4ab+ 2ca3 + 2b2c2 + a2bc3)(2c+ ab2)

= 8abc+ 4(c2a3 + a2b3 + b2c3) + 2abc(a3b+ b3c+ c3a) + a3b3c3

= 8abc+ 4(a2b3 + b2c3 + c2a3) + 2abc(4− a2b2c2) + a3b3c3

= 16abc+ 4(a2b3 + b2c3 + c2a3)− a3b3c3.

Hence, S = P which establishes (2).

Now, for all u, v, w > 0, we have by the Cauchy-Schwarz Inequality that

(u+ v + w)(a2

u+b2

v+c2

w) ≥ (a+ b+ c)2. (3)

Settingu = 2a2 + bc2a, v = 2b2 + ca2b and w = 2c2 + ab2c,

we then have by (1) and (3) that

1 =a2

2a2 + bc2a+

b2

2b2 + ca2b+

c2

2c2 + ab2c≥ (a+ b+ c)2

u+ v + w,

so

(2a2 + bc2a) + (2b2 + ca2b) + (2c2 + ab2c) = u+ v + w ≥ (a+ b+ c)2

from which it follows that

a2 + b2 + c2 + abc(a+ b+ c) ≥ 2(ab+ bc+ ca).


Let a, b, c and d be real numbers with 0 < a < b < c < d. Prove that

b

a+c

b+d

c> 3 + ln

d

a.

There were 14 correct solutions. We present four of them here. Most of the solversapproached the problem along the lines of one of the first two solutions.


SOLUTIONS /413

Solution 1.

Since x > 1 + lnx for x 6= 1,

b

a+c

b+d

c>

Å1 + ln

b

a

ã+(

1 + lnc

b

)+

Å1 + ln

d

c

ã= 3 + ln

d

a

as desired.

Solution 2.

Applying the AM-GM Inequality, we find that the left side is not less than

33

…d

a> 3

Ç1 + ln

3

…d

a

å= 3 + ln

d

a.

Solution 3, by Kee-Wai Lau.

For 0 < a < b < c < d, let

f(a, b, c, d) =b

a+c

b+d

c− ln

d

a,

g(a, b, c) =b

a+c

b+ 1− ln

c

a,

h(a, b) =b

a+ 2− ln

b

a.

An analysis of the partial derivatives reveals that each of its functions strictlyincreases in its final variable, so that

f(a, b, c, d) > f(a, b, c, c) = g(a, b, c) > g(a, b, b) = h(a, b) > h(a, a) = 3,

which yields the desired result.

Solution 4, by the proposers.

Let f(x) = 1/x. A diagram shows that

(b− a)f(a) + (c− b)f(b) + (d− c)f(c) >

∫ d

a

dx

x,

whenceb

a− 1 +

c

b− 1 +

d

c− 1 > ln d− ln a,

as desired.

Editor’s Comments. Two solvers provided a straightforward generalization for anincreasing sequence {ak} of n+ 1 positive reals:

n∑k=1

ak+1

ak> n+ ln

an+1

a1.


414/ SOLUTIONS

4090. Proposed by Nermin Hodzic and Salem Malikic.

Let a, b and c be non-negative real numbers such that a2 + b2 + c2 = 3. Prove that

a

3b2 + 6c− bc+

b

3c2 + 6a− ca+

c

3a2 + 6b− ab≥ 3

8.

We received two correct solutions. We present the solution of the proposers, slightlymodified by the editor.

Using Jensen’s inequality for f(x) = 1x (which is convex on (0,∞)), we have

a

a+ b+ c· 1

3b2 + 6c− bc+

b

a+ b+ c· 1

3c2 + 6a− ca+

c

a+ b+ c· 1

3a2 + 6b− ab

≥Åa(3b2 + 6c− bc)

a+ b+ c+b(3c2 + 6a− ca)

a+ b+ c+c(3a2 + 6b− ab)

a+ b+ c

ã−1,

which we can rearrange to

a

3b2 + 6c− bc+

b

3c2 + 6a− ca+

c

3a2 + 6b− ab≥ (a+ b+ c)2

3(ab2 + bc2 + ca2) + 6(ab+ bc+ ca)− 3abc.

In order to prove the inequality given in the question, it thus suffices to show

(a+ b+ c)2

3(ab2 + bc2 + ca2) + 6(ab+ bc+ ca)− 3abc≥ 3

8,

which holds (by cross multiplying and rearranging) if and only if

8(a+ b+ c)2 ≥ 9(ab2 + bc2 + ca2) + 18(ab+ bc+ ca)− 9abc ⇐⇒8(a2 + b2 + c2) ≥ 9(ab2 + bc2 + ca2) + 2(ab+ bc+ ca)− 9abc.

By the Cauchy-Schwarz inequality, ab+ bc+ ca ≤ a2 + b2 + c2. Note for later thatequality holds if and only if a = b = c = 1. Hence, it suffices to show that

6(a2 + b2 + c2) ≥ 9(ab2 + bc2 + ca2)− 9abc.

Finally, since a2 + b2 + c2 = 3, this reduces to

2 ≥ (ab2 + bc2 + ca2)− abc. (1)

Assume that a ≥ b ≥ c. Then (a− b)(b− c) ≥ 0, equivalent to ab+ bc ≥ b2 + ac.Multiply both sides by a > 0 and rearrange to get abc ≥ ab2 +a2c−a2b. Note thatthe cubic g(b) = 3b − b3 has a local maximum at b = 1, and in fact for all b ≥ 0we have 3b− b3 ≤ g(1) = 2. Hence

abc+ 2 ≥ ab2 + a2c− a2b+ 3b− b3,

which is equivalent toabc+ 2 ≥ ab2 + a2c+ c2b− b(a2 + b2 + c2) + 3b.

Since a2 +b2 +c2 = 3, this shows that abc+2 ≥ ab2 +a2c+c2b, which is equivalentto (1), concluding the proof.


SOLUTIONS /447




Find the greatest positive number k such that

a+ b+ c+ 3k − 3 ≥ kÇ

3

…b

a+ 3

…c

b+ 3

…a

c

åfor any positive numbers a, b and c with abc = 1.

We received three submissions, two of which were correct. We present the solutionof the proposers, modified by the editor.

Using abc = 1 we can rewrite the inequality as

a+ b+ c+ 3(k − 1)− k(3√a2b+

3√b2c+

3√c2a) ≥ 0.

By substituting a = x3, b = y3, and c = z3 (and thus xyz = 1) and defining thehomogeneous cyclic polynomial f : [0,∞)3 → R by

f(x, y, z) = x3 + y3 + z3 + 3(k − 1)xyz − k(x2y + y2z + z2x),

we thus need to find the greatest positive number k such that

f(x, y, z) ≥ 0

for all positive x, y, z ≥ 0 with xyz = 1. We will proceed by first finding thegreatest k such that f(x, y, z) ≥ 0 for all x, y, z ≥ 0 and then showing that this kis optimal for the case of xyz = 1 as well. To accomplish the first part we will usea lemma proven by Pham Kim Hung in his book Secrets in Inequalities (Volume2). As the lemma might be of interest to the reader but is not a common referencewe will give a proof here.

Lemma. Let P (x, y, z) be a cyclic homogeneous polynomial of degree 3. ThenP (x, y, z) ≥ 0 for all x, y, z ≥ 0 if and only if P (1, 1, 1) ≥ 0 and P (u, 1, 0) ≥ 0 forall u ≥ 0.

Proof of the lemma: Clearly the given conditions are necessary. So assume thatP (1, 1, 1) and P (u, 1, 0) are nonnegative for all u ≥ 0. We can write

P (x, y, z) = m∑cyc

x3 + n∑cyc

x2y + p∑cyc

xy2 + qxyz


448/ SOLUTIONS

for some m,n, p, q ∈ R. Then P (1, 1, 1) gives us

3m+ 3n+ 3p+ q ≥ 0,

while setting u = 0 and u = 1 respectively yields m ≥ 0 and 2m+ n+ p ≥ 0. Thederivative of P (x, y, z) in the direction (1, 1, 1) is

3m∑cyc

x2 + n

Å∑cyc

x2 + 2∑cyc

xy

ã+ p

Å∑cyc

x2 + 2∑cyc

xy

ã+ q

∑cyc

xy

= (3m+ n+ p)∑cyc

x2 + (2n+ 2p+ q)∑cyc

xy

Note that3m+ n+ p = m+ (2m+ n+ p) ≥ 0

and(3m+ n+ p) + (2n+ 2p+ q) = 3m+ 3n+ 3p+ q ≥ 0,

thus

(3m+ n+ p)∑cyc

x2 + (2n+ 2p+ q)∑cyc

xy ≥ (3m+ n+ p)

Å∑cyc

x2 −∑cyc

xy

ã≥ 0

by the rearrangement inequality. Since the derivative in the direction (1, 1, 1) isnonnegative for all x, y, z ≥ 0 we only need to show that P (x, y, z) ≥ 0 on theboundary, that is where at least one variable is equal to 0. This now follows fromthe cyclicity of P (x, y, z) and P (a, b, 0) = b3P (ab , 1, 0) for b 6= 0 which finishes theproof of the lemma.

We can now apply the lemma to show that f(x, y, z) ≥ 0 for all x, y, z ≥ 0 if andonly if f(1, 1, 1) ≥ 0 (which clearly holds), and f(u, 1, 0) ≥ 0. Defining

g(u) = f(u, 1, 0) = u3 − ku2 + 1,

we obtain g′(u) = u(3u − 2k). Thus the minimum of g(u) occurs at u = 2k3 and

g(u) is positive for all u ≥ 0 if and only if g( 2k3 ) ≥ 0, which occurs for k ≤ 3

3√4.

It remains to be shown that this is best possible for the case of xyz = 1 as well.Suppose k > 3

3√4and set s = 2k

3 . Consider h(w) = f(s, 1, w). Since h(w) is

continuous and h(0) = g(s) < 0 there exists a positive t such that h(t) < 0. Set

x =s

3√st, y =

13√st

and z =t

3√st.

Then x, y, z ≥ 0 with xyz = 1 and

f(x, y, z) = stf(s, 1, t) = sth(t) < 0.

It follows that the answer to the question is k = 33√4

.

Crux Mathematicorum, Vol. 42(10), December 2016

SOLUTIONS /449


Show thatña2 + 16a+ 80

16 (a+ 4)+

2√2 (b2 + 16)

ô ñb2 + 16b+ 80

16 (b+ 4)+

2√2 (a2 + 16)

ô≥ 9

4

for all a, b > 0. When does equality hold?

We received ten correct submissions. We present two solutions.


Sincea2 + 16a+ 80

16 (a+ 4)= 1 +

a2 + 42

16 (a+ 4),

we haveÇa2 + 16a+ 80

16 (a+ 4)+

2√2 (b2 + 16)

åÇb2 + 16b+ 80

16(b+ 4)+

2√2 (a2 + 16)

å=

Ç1 +

a2 + 42

16 (a+ 4)+

2√2 (b2 + 42)

åÇ1 +

b2 + 42

16 (b+ 4)+

2√2 (a2 + 42)

åand, combining Cauchy-Schwarz Inequality and inequality

√2 (u2 + v2) ≥ u + v,

we obtainÇ1 +

a2 + 42

16 (a+ 4)+

2√2 (b2 + 42)

åÇ1 +

2√2 (a2 + 42)

+b2 + 42

16 (b+ 4)

å≥

(1 · 1 +

a2 + 42

16 (a+ 4)·

2√2 (a2 + 42)

+

2√

2 (b2 + 42)·

b2 + 42

16 (b+ 4)

)2

=

(1 +

1

4

√2 (a2 + 42)

a+ 4+

1

4

√2 (b2 + 42)

b+ 4

)2

≥Å

1 +1

4+

1

4

ã2=

9

4.

Since in inequality√

2 (u2 + v2) ≥ u+ v equality occurs if and only if u = v, itis easy to see that the equality holds if and only if a = b = 4.

Solution 2, by AN-anduud Problem Solving Group.

Using AM-GM inequality, we get»2(b2 + 16) =

…8 · b

2 + 16

4≤ 1

2

Å8 +

b2 + 16

4

ã=b2 + 48

8. (1)

Applying AM-GM inequality and using (1), we have


450/ SOLUTIONS

a2 + 16a+ 80

16(a+ 4)+

2√2(b2 + 16)

≥ a2 + 16a+ 80

16(a+ 4)+

16

b2 + 48

=(a2 + 48) + (a+ 4)2 + 24(a+ 4)

32(a+ 4)+

16

b2 + 48

=a2 + 48

32(a+ 4)+a+ 4

32+

1

4+

1

4+

1

4+

16

b2 + 48

≥ 6 6

a2 + 48

32(a+ 4)· a+ 4

32· 1

4· 1

4· 1

4· 16

b2 + 48

=3

2· 6

a2 + 48

b2 + 48. (2)

Similarly,b2 + 16b+ 80

16(b+ 4)+

2√2(a2 + 16)

≥ 3

2· 6

b2 + 48

a2 + 48. (3)

Multiplying (2) and (3), we obtain the desired inequality. Equality holds onlywhen a = b = 4.

4093. Proposed by Dragolijub Milosevic.

Let ABC be an arbitrary triangle. Let r and R be the inradius and the circum-radius of ABC, respectively. Let ma be the length of the median from vertex Ato side BC and let wa be the length of the internal bisector of ∠A to side BC.Define mb,mc, wb and wc similarly. Prove that

a2

mawa+

b2

mbwb+

c2

mcwc≤ 4

ÅR

r− 1

ã.

We received five correct solutions and present the solution by Andrea Fanchini.

We have that4

ÅR

r− 1

ã=abcs− 4K2

K2

where s is the semiperimeter and K is the area of the triangle. We know (seeWei-Dong Jiang and Mihaly Bencze, JMI Volume 5 Number 3, 2011, p. 365) thatmawa ≥ s(s− a), so we have to prove

a2

s(s− a)+

b2

s(s− b)+

c2

s(s− c)≤ abcs− 4K2

K2,

that is

a2(s−b)(s−c)+b2(s−a)(s−c)+c2(s−a)(s−b) ≤ s (abc− 4(s− a)(s− b)(s− c)) .

In fact, the inequality can be replaced by equality since both sides equal:

s (abc− 4(s− a)(s− b)(s− c))

=1

4

(a4 + b4 + c4 + 2a2bc+ 2ab2c+ 2abc2 − 2a2b2 − 2b2c2 − 2c2a2

).


SOLUTIONS /451


Let x1, x2, . . . , xn be real numbers such that 0 ≤ x1 ≤ x2 ≤ · · · ≤ xn. Prove that

n− 1 + cosh

(n∑k=1

(−1)k−1xk

)≤

n∑k=1

coshxk ≤ n− 1 + cosh

(n∑k=1

xk

).

We received four correct and complete solutions. We present the solution by theproposer.

Let f be the function defined on R by f(x) = (coshx)− 1. As a lemma, we showthat

f(a) + f(b) ≤ f(a+ b) if ab ≥ 0 and f(a) + f(b) ≥ f(a+ b) if ab ≤ 0.

The result clearly holds if b = 0. Fix b > 0 and consider the function φ : x 7→f(x + b) − f(x). This function φ is differentiable and its derivative, defined byφ′(x) = f ′(x+ b)− f ′(x), is nonnegative (since f ′ = sinh is nondecreasing). Thusφ(a) ≥ φ(0) = f(b) if a ≥ 0 and φ(a) ≤ f(b) if a ≤ 0. The result follows whenb > 0. In a similar way, we see that it also holds when b < 0.

This lemma and an easy induction lead to

f(x1) + f(x2) + · · ·+ f(xn) ≤ f(x1 + x2 + · · ·+ xn)

if x1, x2, . . . , xn ≥ 0. The right-hand inequality immediately follows.

We also prove the left-hand inequality by induction. The case n = 1 is obvious.Consider the case n = 2: Since x1(−x2) ≤ 0, we have

f(x1) + f(−x2) ≥ f(x1 − x2) or cosh(x1) + cosh(x2)− cosh(x1 − x2) ≥ 1,

the desired inequality. Now, assume that for some n ≥ 2(n∑k=1

coshxk

)≥ n− 1 + cosh

(n∑k=1

(−1)k−1xk

)(1)

whenever 0 ≤ x1 ≤ x2 ≤ · · · ≤ xn.

Let x1, x2, . . . , xn, xn+1 be such that 0 ≤ x1 ≤ x2 ≤ · · · ≤ xn ≤ xn+1. Then weobserve that if n is even, then

xn+1(x1 − x2 + · · ·+ xn−1 − xn) ≤ 0

and if n is odd, then

(−xn+1)(x1 − x2 + · · · − xn−1 + xn) ≤ 0.

Applying the lemma and using cosh(−x) = cosh(x), we obtain

coshxn+1 − 1 + cosh

(n∑k=1

(−1)k−1xk

)− 1 ≥ cosh

(n+1∑k=1

(−1)k−1xk

)− 1


452/ SOLUTIONS

in either case. With the help of (1), we are led to(n+1∑k=1

coshxk

)≥ coshxn+1 + n− 1 + cosh

(n∑k=1

(−1)k−1xk

)

≥ n+ cosh

(n+1∑k=1

(−1)k−1xk

).

This completes the induction step and the proof.


Let a, b and c be positive real numbers with 1a + 1

b + 1c = 3. Prove that

ab(a+ b) + bc(b+ c) + ac(a+ c) ≥ 2

3(a2 + b2 + c2) + 4abc.

There were 14 correct solutions. We present six different ones here.


With 1 = ax = by = cz and x+ y + z = 3, the inequality is equivalent to

(x+ y + z)[z2(x+ y) + x2(y + z) + y2(z + x)]

≥ 2(x2y2 + y2z2 + z2x2) + 4(x+ y + z)(xyz)

= 2(xy + yz + zx)2.

The difference between the two sides is

(x+ y + z)[(x+ y + z)(xy + yz + zx)− 3xyz]− 2(xy + yz + zx)2

= (x+ y + z)2(xy + yz + zx)− 2(xy + yz + zx)2 − 3xyz(x+ y + z)

≥ 3(xy + yz + zx)(xy + yz + zx)− 2(xy + yz + zx)2 − 3xyz(x+ y + z)

= (xy + yz + zx)2 − 3xyz(x+ y + z) ≥ 0

(from two applications of the inequality (u+v+w)2 ≥ 3(uv+vw+wu)). Equalityoccurs iff 1 = x = y = z = a = b = c.

Solution 2, by Andrew Siefker and Digby Smith (done independently).

When a, b, c, x, y, z are all positive, we have

(a2yz + b2zx+ c2xy)(x+ y + z)− (a+ b+ c)2xyz

= z[(a2y2 + b2x2 − 2abxy] + y[a2z2 + c2x2 − 2acxz] + x[b2z2 + c2y2 − 2bcyz] ≥ 0,

so thata2

x+b2

y+c2

z≥ (a+ b+ c)2

x+ y + z.

Equality occurs iff a : b : c = x : y : z. (This also results from Cauchy’s Inequality.)


SOLUTIONS /453

Let bx = cy = az = 1. Then (since ab+ bc+ ca = 3abc),

a2b+ b2c+ c2a ≥ abc(a+ b+ c)2

ab+ bc+ ca=

(a+ b+ c)2

3=a2 + b2 + c2

3+ 2abc.

Let cx = ay = bz = 1. Then,

a2c+ b2a+ c2b =(a+ b+ c)2

3=a2 + b2 + c2

3+ 2abc.

Adding these two inequalities yields the result. Equality occurs iff a = b = c = 1.

Solution 3, by Prithwijit De.

Since ab+ bc+ ca = 3abc, it follows that

2(a2 + b2 + c2) = (b− c)2 + (c− a)2 + (a− b)2 + 6abc.

Also,

ab(a+ b) + bc(b+ c) + ac(a+ c) = a(b− c)2 + b(c− a)2 + c(a− b)2 + 6abc

anda

Å1

b+

1

c

ã= a

Å3− 1

a

ã= 3a− 1,

b

Å1

c+

1

a

ã= 3b− 1 and c

Å1

a+

1

b

ã= 3c− 1.

Therefore,

3[ab(a+ b) + bc(c+ a) + ca(a+ c)]− 2(a2 + b2 + c2)− 12abc

= 3[a(b− c)2 + b(c− a)2 + c(a− b)2] + 18abc

− [(b− c)2 + (c− a)2 + (a− b)2 + 6abc]− 12abc

= (3a− 1)(b− c)2 + (3b− 1)(c− a)2 + (3c− 1)(a− b)2

= a

Å1

b+

1

c

ã(b− c)2 + b

Å1

c+

1

a

ã(c− a)2 + c

Å1

a+

1

b

ã(a− b)2

≥ 0,

with equality iff a = b = c = 1. The desired result follows.


Replacing the 3 in the denominator by (ab + bc + ca)/abc and multiplying byab+ bc+ ca, we obtain the equivalent homogeneous inequality

a2b2(a+ b) + b2c2(b+ c) + c2a2(c+ a) + abc[(a+ b)2 + (b+ c)2 + (c+ a)2]

≥ 2(a3bc+ b3ca+ c3ab) + 4(ab2c2 + bc2a2 + ca2b2)

which in turn is equivalent to


454/ SOLUTIONS

a3b2 + a2b3 + b3c2 + b2c3 + c3a2 + c2a3 ≥ 2(ab2c2 + bc2a2 + ca2b2).

By the arithmetic-geometric means inequality, we have that

2a3b2 + b2c3 ≥ 3ca2b2; 2b3c2 + c2a3 ≥ 3ab2c2; 2c3a2 + a2b3 ≥ 3bc2a2;

a3b2 + 2b2c3 ≥ 3ab2c2; b3c2 + 2c2a3 ≥ 3bc2a2; c3a2 + 2a2b3 ≥ 3ca2b2.

Adding these inequalities yields the desired result, with equality iff a = b = c = 1.

Solution 5, by AN-Anduud Problem Solving Group; and Dionne Bailey, ElsieCampbell and Charles R. Diminnie (independently).

Since a2 + b2 + c2 ≥ ab+ bc+ ca = 3abc and x+ (1/x) ≥ 2 for x > 0, we have that

3[ab(a+ b) + bc(b+ c) + ca(c+ a)]

=

Å1

a+

1

b+

1

c

ã[ab(a+ b) + bc(b+ c) + ca(c+ a)]

= 2(a2 + b2 + c2) + 2(ab+ bc+ ca) +bc

a(b+ c) +

ca

b(c+ a) +

ab

c(a+ b)

= 2(a2 + b2 + c2) + 6abc+ a2Åb

c+c

b

ã+ b2

( ca

+a

c

)+ c2

Åa

b+b

a

ã≥ 6abc+ 6abc+ 2(a2 + b2 + c2),

yielding the desired result. Equality holds iff a = b = c = 1.


Applying the arithmetic-harmonic means inequality to the three pairs (1/a, 1/b),(1/b, 1/c) and (1/c, 1/a), we find that

3 =1

a+

1

b+

1

c≥ 2

a+ b+

2

b+ c+

2

c+ a.

This is equivalent to

3(a+ b)(b+ c)(c+ a) ≥ 2[(b+ c)(c+ a) + (c+ a)(a+ b) + (a+ b)(b+ c)] ⇐⇒6abc+ 3[ab(a+ b) + bc(b+ c) + ca(c+ a)] ≥ 2(a2 + b2 + c2) + 6(ab+ bc+ ca)

= 2(a2 + b2 + c2) + 18abc ⇐⇒

ab(a+ b) + bc(b+ c) + ca(c+ a) ≥ 2

3(a2 + b2 + c2) + 4abc.

Equality holds iff a = b = c = 1.

Editor’s Comments. Four solvers reformulated the inequality as in Solution 4 andapplied Muirhead’s Inequality [3, 2, 0] ≥ [2, 2, 1], where

[p, q, r] = apbqcr + apbrcq + aqbpcr + aqbrcp + arbpcq + arbqcp.

Students Ahmad Talafha and Kevin Wunderlich gave a variant of Solution 2.


SOLUTIONS /455

4096. Proposed by Abdilkadir Altintas.

Let ABC be a heptagonal triangle with BC = a, AC = b and AB = c. SupposeCN is the internal angle bisector of ∠BCA, BM is the median of triangle ABCand K is the symmedian point of ABC. Show that N,K and M are collinear.

We received six correct submissions. We present a combined solution based onthose received from Michel Bataille and Titu Zvonaru.

In barycentric coordinates with respect to the vertices A,B,C of the triangle wehave N = (a : b : 0), K = (a2 : b2 : c2), M = (1 : 0 : 1). The points N , K andM are collinear if and only if ∣∣∣∣∣∣

a a2 1b b2 00 c2 1

∣∣∣∣∣∣ = 0,

which is equivalent toc2 + ab = a2. (1)

Consider the cyclic quadrilateral ABEC. Since ADCEFGB is a regular heptagon,|CE| = |AB| = c and |EB| = |EA| = |BC| = a. Applying Ptolemy’s theorem toquadrilateral ABEC we have

|AB| · |CE|+ |AC| · |BE| = |BC| · |EA|,

which gives us (1). Therefore the points N , K and M are collinear.


Let ai, 1 ≤ i ≤ 6 be real numbers such that

6∑i=1

ai =15

2and

6∑i=1

a2i =45

4.

Prove that∏6i=1 ai ≤

5

2.


456/ SOLUTIONS

We received five submissions, four of which were incorrect for various reasons. Wepresent the proposer’s solution, modified by the editor.

Using Jensen’s inequality with the function g(x) = x2 we have

5

(5∑i=1

a2i

)≥

(5∑i=1

ai

)2

and thus using the assumptions from the question,

5

Å45

4− a26ã≥Å

15

2− a6ã2,

which lets us conclude 0 ≤ a6 ≤ 52 . By symmetry we obtain 0 ≤ ai ≤ 5

2 for

i = 1, . . . , 6. If any of the six variables is zero, then∏6i=1 ai = 0 ≤ 5

2 , and we aredone. Thus we assume that all ai are positive and at most 5

2 .

Note that∑1≤i<j≤6

aiaj =1

2

[Å 6∑i=1

ai

ã2−

6∑i=1

a2i

]=

1

2

ñÅ15

2

ã2− 45

4

ô=

45

2.

Define the polynomial P (x) =∏6i=1(x− ai) which can be written as

P (x) = x6 − 15

2x5 +

45

2x4 −mx3 + nx2 − qx+ p

for p =∏6i=1 ai and suitable m,n, q ∈ R. Set f(x) = P (x)

x and define the sequenceof polynomials

P (x) = P0(x), P1(x), P2(x), P3(x) = Q(x),

where Pi(x) = xi+1f (i)(x) is the numerator of the i-th derivative of f(x). Notethat we can calculate

Pi+1(x) = xi+2f (i+1)(x) = xi+2 d

dx

Pi(x)

xi+1= xP ′i (x)− (i+ 1)Pi(x).

Now consider the positive roots of Pi+1(x). If Pi(x) has a root at α with mul-tiplicity k, then P ′i (x) has a root at α with multiplicity k − 1, and therefore sodoes Pi+1(x). On the other hand suppose we have two distinct positive roots ofPi(x). Then they are also roots of f (i)(x). By Rolle’s theorem (note that f (i)(x)is continuous for x > 0), there exists a root of f (i+1)(x) (and thus of Pi+1(x))between these two roots. Using those two facts we can conclude that if Pi(x) hasn positive roots then Pi+1(x) has at least n − 1 positive roots. As P (x) has sixpositive roots, Q(x) must have at least three positive roots.

We can calculateQ(x) = 60x6 − 180x5 + 135x4 − 6p

and from thereQ′(x) = 180x3(x− 1)(2x− 3).


SOLUTIONS /457

If Q(x) had two roots in the interval (0, 1), then Q′(x) would have a root in theinterval (0, 1) (by Rolle’s theorem and our remarks about roots with multiplicity),which is not the case. Similarly Q(x) cannot have more than one root greater than32 . Since Q(x) has at least three positive roots, though, it has a root in the interval[1, 32 ]. Looking at Q′(x) we can see that Q(x) is decreasing on this interval and weobtain Q(1) ≥ 0 and therefore p ≤ 5

2 , which finishes the proof.

The bound can be obtained by setting one variable to 52 and the others to 1.

4098. Proposed by Ardak Mirzakhmedov.

Let α, β and γ be acute angles such that α+ β = γ. Show that

cosα+ cosβ + cos γ − 1 ≥ 2√

cosα · cosβ · cos γ.

We received six correct submissions. We present the solution by Arkady Alt.

Note first that for all a, b, c, d ∈ R, we have (ac−bd)2−(ad−bc)2 = (a2−b2)(c2−d2),so (a2 − b2)(c2 − d2) ≤ (ac− bd)2. In particular, if a > b > 0 and c > d > 0, then

ac− bd ≥√a2 − b2 ·

√c2 − d2. (1)

Next,cosα+ cosβ + cos γ − 1 = 2 cos

α+ β

2· cos

α− β2− 2 sin2 γ

2

= 2

Åcos

γ

2· cos

α− β2− sin

γ

2· sin α+ β

2

ã. (2)

Since α+β2 = γ

2 ∈ (0, π4 ) we have

cosα− β

2> cos

α+ β

2= cos

γ

2> sin

γ

2.

Hence, if we leta = cos

γ

2, b = d = sin

γ

2and c = cos

α− β2

,

then a > b > 0 and c > d > 0 so applying (1) we obtain

cosγ

2· cos

α− β2− sin

γ

2· sin α+ β

2≥…

cos2γ

2− sin2 γ

2·…

cos2α− β

2− sin2 α+ β

2

=√

cos γ

…1

2(1 + cos(α− β)− (1− cos(α+ β))

=√

cos γ√

cosα · cosβ

=√

cosα · cosβ · cos γ. (3)

Substituting (3) into (2), we then have

cosα+ cosβ + cos γ − 1 ≥ 2√

cosα · cosβ · cos γ,

thus completing the proof.


458/ SOLUTIONS

4099. Proposed by Lorian Saceanu.

Let ABC be an acute angle triangle. Suppose the internal bisectors of angles A,Band C intersect the sides of ABC in points A′, B′ and C ′ and they intersect thecircumcircle of ABC in points L,M and N respectively. Let I be the point ofintersection of all internal bisectors. Show that:

a)AI

IL=IA′

A′L,

b)

…AI

IL+

…BI

IM+

…CI

IN≥ 3.

We received seven submissions, all correct, and will feature parts from two ofthem.

Solution to part a), by Prithwimit De and B.J. Venkatachala (together).

In triangle IBL, ∠IBL = ∠BIL = A+B2 . Thus, IL = BL = 2R sin A

2 , where R isthe circumradius of ∆ABC. Moreover, since AI = r

sin(A/2) , where r is the inradius

of ∆ABC,AI

IL=

r

BL sin A2

.

Observe that

IA′

A′L=

[BIA′]

[BLA′]=BI sin B

2

BL sin A2

=r

BL sin A2

=AI

IL.

This settles part a).

Solution to part b), by Salem Malikic, modified by the editor.

We first transform the form of AIIL obtained above into something more helpful. Inparticular, we use area formulas

r =[ABC]

s=bc sin2A

2s,

the sine law 2R =a

sinA, and the half-angle formulas together with the cosine law

sin2A

sin2 A2

= 4 cos2A

2=

2s(b+ c− a)

bc

in turn to get

AI

IL=

r

BL sin A2

=r

2R sin2 A2

=bc sin2A

2sa sin2 A2

=b+ c− a

a.

This, with analogous expressions for BIIM and CI

IN , reduces the inequality of partb) to …

b+ c− aa

+

…c+ a− b

b+

…a+ b− c

c≥ 3, (1)


SOLUTIONS /459

which is an inequality due to Sorin Radulescu that holds for all acute triangles.[Editor’s comment. Malikic found a proof on the web pagewww.artofproblemsolving.com/community/c6h1102804p5008234,which he reproduced while adding some helpful details as follows.] After squaringboth sides of (1) we need to prove that

∑cyclic

b+ c− aa

+ 4∑cyclic

(a+ b− c)(a+ c− b)

4bc≥ 9,

which is ∑cyclic

b+ c− aa

+ 4

Åsin

A

2+ sin

B

2+ sin

C

2

ã≥ 9.

Since cosine is a concave function on [0, π2 ], Popoviciu’s inequality tells us that

cosA+ cosB + cosC + 3 cos

ÅA+B + C

3

ã≤ 2

Åcos

A+B

2+ cos

B + C

2+ cos

C +A

2

ã,

or equivalently,

2

Åsin

A

2+ sin

B

2+ sin

C

2

ã≥∑cyclic

b2 + c2 − a2

2bc+

3

2. (2)

With the help of inequality (2) it remains to prove that∑cyclic

b+ c− aa

+∑cyclic

b2 + c2 − a2

bc≥ 6.

Setting a = y + z, b = x + z and c = x + y reduces the last inequality to theequivalent ∑

cyclic

(x3 − x2y − x2z + xyz) ≥ 0,

which directly follows from Schur’s inequality, thus completing the proof. More-over, from the last step we see that for equality one must have x = y = z, whichimplies that ∆ABC must be equilateral. It is easily seen that equality is indeedachieved for an equilateral triangle.

Editor’s Comments. Barroso Campos observed that the inequality of part b) mightfail when ∆ABC has an obtuse angle. One can easily construct counterexamplesusing the left-hand-side of (1); for example, with sides a = 9, b = 10, c = 18 we geta sum slightly smaller than 2.993.


Let ABC be an arbitrary triangle with area S, ∠A < 90◦ and sides BC = a,AC = b and AB = c. Show that

c cosB

ac+ 2S+

b cosC

ab+ 2S<

a

2S.


460/ SOLUTIONS

We received nine submissions, of which eight were correct. We present the solutionby Salem Malikic, slightly modified by the editor.

Using the area formula S = 12ac sinB we get

c cosB

ac+ 2S=

c cosB

ac+ ac sinB=

cosB

a(1 + sinB);

treating the second term similarly, the left hand side of the inequality can bere-written as 1

a·Å

cosB

1 + sinB+

cosC

1 + sinC

ã.

On the right hand side of the given inequality we have

a

2S=

a

ac sinB(area formula)

=a

a · a sinCsinA · sinB

(sine law)

=sin(B + C)

a sinC sinB(A+B + C = π, so sin(A) = sin(B + C))

=1

a·Å

cosC

sinC+

cosB

sinB

ã(sum of angles formula for sine)

Hence, since a > 0, the given inequality is equivalent to

cosB

1 + sinB+

cosC

1 + sinC<

cosB

sinB+

cosC

sinC,

and, by moving all the terms to one side, this is in turn equivalent to

cosB

sinB(1 + sinB)+

cosC

sinC(1 + sinC)> 0. (†)

If neither of ∠B and ∠C is obtuse then all the trigonometric functions are non-negative and the last inequality clearly holds (note that both terms cannot besimultaneously zero, as that would imply ∠B = ∠C = 90◦).

Assume now that one of the angles B and C, say without loss of generality C, isobtuse. As cos(C) = − cos(π − C) and sin(C) = sin(π − C), the inequality (†)holds if and only if

cosB

sinB(1 + sinB)>

cos(π − C)

sin(π − C)(1 + sin(π − C)).

However, A + B + C = π and C obtuse imply 0 < B < π − C < π2 ; since on

the interval (0, π/2) cosine is decreasing and sine is increasing, this last inequalityfollows, concluding the proof of (†) and thus also of the desired inequality.


solutions...show that for any triangle abc, the following inequality holds sinasinbsinc † 1 sina+...

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