SOLUTIONS - Canadian Mathematical Society · 356/ Solutions We received 6 submissions, all of which...

350/ Solutions

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

Statements of the problems in this section originally appear in 2018: 44(8), p. 340–343;44(9), p.387–390; 44(10): 423–426.

4376. Proposed by Marius Dragan and Neculai Stanciu.

Let A and B be two matrices in Mn(C) such that AB = −BA. Prove that

det(A4 +A2B2 + 2A2 + In) ≥ 0.

We received 6 solutions and will feature just one of them here. We present thesolution by Missouri State University Problem Solving Group.

The result is false as stated. Let

A =

[α 00 −α

]and B =

[0 10 0

],

where α = (1 + i)/√

2. Then AB = −BA, but

det(A4 +A2B2 + 2A2 + In

)=(1 + α2

)4= −4.

We assume that the intended hypothesis was A,B ∈Mn(R). Note that(In +A2 +AB

) (In +A2 +BA

)= In +A2 +BA+A2 +A4 +A2BA+AB +ABA2 +ABBA

= In +A2 −AB +A2 +A4 +A2BA+AB −A2BA−ABAB= In +A2 −AB +A2 +A4 +A2BA+AB −A2BA+A2B2

= A4 +A2B2 + 2A2 + In.

Now

det(In +A2 +AB

)= det (In +A (A+B))

= det (In + (A+B)A)

= det(In +A2 +BA

),

using Sylvester’s Theorem that det(In +XY ) = det(In + Y X). Finally, we have

det(A4 +A2B2 + 2A2 + In

)= det

((In +A2 +AB

) (In +A2 +BA

))= det

(In +A2 +AB

)det(In +A2 +BA

)=

(det(In +A2 +AB

))2≥ 0.

Crux Mathematicorum, Vol. 45(6), July 2019

Solutions /351

4377. Proposed by Tidor Vlad Pricopie and Leonard Giugiuc.

Let x ≥ y ≥ z > 0 such that x+ y + z + xy + xz + yz = 1 + xyz. Find minx.

We received 7 correct solutions and 1 incorrect submission.We feature the solutionbased on the approach of Ramanujan Srihari and the collaboration between LeonardGiugiuc and Tidor Vlad Pricopie, done independently.

We first show that xy+xz+yz 6= 1. Otherwise, we would have that x+y+z = xyzand

z = z(xy + xz + yz) = x+ y + z + z2(x+ y),

leading to

0 = (x+ y)(1 + z2),

which contradicts the hypotheses.

Let x = tanu, y = tan v, z = tanw where π/2 > u ≥ v ≥ w > 0. Then

tan(u+ v + w) =x+ y + z − xyz1− xy − xz − yz = 1.

Therefore

u+ v + w = π/4 or u+ v + w = 5π/4.

In the latter case, we would have u > 5π/12 and x > 1.

When u+ v + w = π/4, then u ≥ π/12. From the formula for tan 3θ, we see thattanπ/12 satisfies the equation

0 = t3 − 3t2 − 3t+ 1 = (t+ 1)(t2 − 4t+ 1),

so that x ≥ 2 −√

3. However, it is possible that u = v = w = π/12, so that theequation is satisfied by x = y = z = 2−

√3.

Therefore, for all positive solutions of the equation, x ≥ 2 −√

3, and the lowerbound is attained.

4378. Proposed by Tarit Goswami.

Find all k such that the following limit exists

limn→∞

{k · Fn+1 −n∑i=0

τ i},

where Fn is the nth Fibonacci number and τ is the golden ratio.

Five correct solutions and three incorrect or incomplete solutions were receivedto the problem when the brackets are used to simply display the summand. Thefirst solution by Sushanth Sathish Kumar solves this problem. However, the Prob-lem Solving Group at Missouri State University took the brackets to refer to thefractional part. Theirs is the second solution below.

Copyright © Canadian Mathematical Society, 2019

352/ Solutions

Solution 1, by Sushanth Sathish Kumar.

Using Binet’s formula for Fn+1 and the relation 1− τ = −1/τ , we have that

kFn+1 −n∑i=0

τ i =k√5

(τn+1 − (1− τ)n+1)− τn+1 − 1

τ − 1

=k√5

(τn+1 − (1− τ)n+1)− τn+2 + τ

= τ +

(k√5− τ)τn+1 − k√

5(1− τ)n+1.

Since limn→∞ τn+1 = ∞ and limn→∞(1 − τ)n+1 = 0, the limit can exist if andonly if k =

√5τ = (5 +

√5)/2, in which case the limit is τ .

Solution 2, by Missouri State University Problem Solving Group.

Let

xn = kFn+1 −n∑i=0

τ i

and let {xn} = xn − bxnc be the fractional part of xn. We prove that the limit of{xn} exists if and only if k = u+ vτ for some integers u and v.

Suppose first that k = u+ vτ . Since τn = τFn +Fn−1 for each positive integer n,

xn = kFn+1 − (τn+2 − τ) = (k − 1)Fn+1 − τFn+2 + τ. (1)

By Binet’s formula, limn→∞ τFn − Fn+1 = 0, so that

xn = (k − 1)Fn+1 − τFn+2 + τ

= (u− 1)Fn+1 + vτFn+1 − τFn+2 + τ

= (u− 1)Fn+1 + vFn+2 − Fn+3 + 1 + (τ − 1 + εn)

where limn→∞ εn = 0. Hence, for sufficiently large n, {xn} = τ − 1 + εn and solimn→∞{xn} = τ − 1.

Now, suppose that lim{xn} = L and let

yn = bxnc − 1 = xn − {xn} − 1.

Then, for n ≥ 3, from (1), we have that

yn − yn−1 − yn−2 = (xn − xn−1 − xn−2)− ({xn} − {xn−1})− {xn−2}) + 1

= −τ + ({xn−1}+ {xn−2} − {xn}) + 1.

Since {yn−yn−1−yn−2} is a convergent sequence of integers with limit L−(τ−1),it is eventually constant. Since 0 ≤ L ≤ 1 and 0 < τ − 1 < 0, the limit must be 0and so yn = yn−1 + yn−2 for n ≥ N . Therefore, for n ≥ N ,

yn = aFn+1−N + bFn−N


Solutions /353

for n ≥ N , where a = yN and b = yN−1.

From (1),

k − 1 =xn + τFn+2 − τ

Fn+1

=yn + {xn}+ 1 + τFn+2 − τ

Fn+1

= aFn+1−N

Fn+1+ b

Fn−NFn+1

+ τFn+2

Fn+1+{xn} − (τ − 1)

Fn+1.

Let n tend to infinity. Then

k − 1 = aτ−N + bτ−(N+1) + τ2

= a(τ − 1)N + b(τ − 1)N+1 + τ2,

so that k = u+ vτ for some integers u and v.

4379. Proposed by Kadir Altintas and Leonard Giugiuc.

Let triangle ABC share its vertices with three vertices of a regular heptagon; inparticular, let B coincide with vertex 1, C with vertex 2, and A with vertex 4.Let I be the incenter and let G be the centroid of ABC, respectively. Suppose BIintersects AC in D and CI intersects AB in E. Show that the points D,G and Eare collinear.

We received 5 solutions. Presented is the one by Andrea Fanchini, lightly edited.

We use barycentric coordinates with reference to the triangle ABC, with sidelengths a, b, and c. The vertices of the triangle have coordinates A(a : 0 : 0),B(0 : b : 0) and C(0 : 0 : c), the incentre has coordinates I(a : b : c) and thecentroid G(1 : 1 : 1). Using that D is on both lines BI and AC, we obtain thecoordinates D(a : 0 : c). Similarly we get E(a : b : 0). The points G, D and E arecollinear if and only if

∣∣∣∣∣∣∣∣∣∣1 1 1

a 0 c

a b 0

∣∣∣∣∣∣∣∣∣∣= ac− bc+ ab = 0.

Now we know that the side lengths of the heptagonal triangle satisfy the relations

bc = c2 − a2, ac = b2 − a2, ab = c2 − b2.

After substituting these relations, we are done.


354/ Solutions

4380. Proposed by George Apostolopoulos.

Let a, b and c be the side lengths of a triangle ABC with inradius r and circum-radius R. Prove that

a2 tanA

2+ b2 tan

B

2+ c2 tan

C

2≤ 3√

3R3(R− r)2r2

.

We received 12 submissions, including the one from the proposer, all correct. Wepresent two solutions, the second one of which gives a sharper inequality.

Solution 1, by Kee-Wai Lau.

Let S denote the semiperimeter of triangle ABC. The following identities andinequalities are all well known:

a

sinA=

b

sinB=

c

sinC= 2R (1)

sinA cosA+ sinB cosB + sinC cosC =rS

R2(2)

R ≥ 2r (Euler’s Inequality) (3)

s ≤ 3√

3R

2(4)

By (1) we have

a2 tanA

2= 4R2(sin 2A)(tan

A

2)

= 4R2(sinA)(2 sinA

2cos

A

2)(tan

A

2)

= 8R2(sinA)(sin 2A

2)

= 4R2(sinA)(1− cosA).

Similarly, b2 tan B2 = 4R2(sinB)(1− cosB) and c2 tan C

2 = 4R2(sinC)(1− cosC).Hence, by (1), (2), (3), and (4) we have∑

cyc

a2 tanA

2= 4R2

(∑cyc

sinA−∑cyc

sinA cosA)

= 4R2(a+ b+ c

2R− rs

R2

)= 4s(R− r)≤ 6√

3R(R− r)

=3√

3R(2r)2(R− r)2r2

≤ 3√

3R3(R− r)2r2

and we are done.


Solutions /355

Solution 2, by Arkady Alt.

We prove the inequality that∑cyc

a2 tanA

2≤ 6√

3R(R− r)

which is sharper than the proposed result since

6√

3R(R− r) ≤ 3√

3R3(R− r)2r2

⇐⇒ 2r ≤ R

which is Euler’s Inequality.

Using the known results that

tanA

2=

r

s− a,∑cyc

a

s− a =4R− 2r

rand s ≤ 3

√3R

2,

we obtain ∑cyc

a2 tanA

2≤ 6√

3R(R− r) ⇐⇒

∑cyc

a2

s− a ≤6√

3R(R− r)r

⇐⇒

∑cyc

( a2

s− a + a)≤ 6√

3R(R− r)r

+ 2s ⇐⇒

∑cyc

( sa

s− a)≤ 6√

3R(R− r)r

+ 2s ⇐⇒

s ·(4R− 2r

r

)≤ 6√

3R(R− r)r

+ 2s ⇐⇒

s(2R− r) ≤ 3√

3R(R− r) + sr ⇐⇒2s(R− r) ≤ 3

√3R(R− r) ⇐⇒

s ≤ 3√

3R

2

and the proof is complete.

4381. Proposed by Mihaela Berindeanu.

Let ABC be an acute triangle with circumcircle Γ1 and circumcenter O. Supposethe open ray AO intersects Γ1 at point D and E is the middle point of BC. Theperpendicular bisector of BE intersects BD in P and the perpendicular bisectorof EC intersects CD in Q. Finally suppose that circle Γ2 with center P and radiusPE intersects the circle Γ3 with center Q and radius QE in X. Prove that AX isa symmedian in M ABC.


356/ Solutions

We received 6 submissions, all of which were correct, and feature the solution byK.V. Sudharshan.

Define X ′ to be the intersection of the A-symmedian with Γ1. We shall prove thatX ′ = X. Let AE intersect Γ1 at a point Y . Observe that since AX ′, AY areisogonal (by definition), we have ∠BAX ′ = ∠Y AC, and so BX ′ = CY . Since Eis the midpoint of BC, we see that BX ′E and CY E are congruent triangles (byside-angle-side). Consequently,

∠EX ′B = ∠CY E = ∠CY A = ∠CBA.

Also, since AD is a diameter of Γ1 and P ∈ BD, we have PB ⊥ AB, so that∠CBA = 90◦ − ∠PBE. Thus (because P is on the perpendicular bisector of thesegment BE),

∠EPB = 180◦ − 2 · ∠PBE = 2 · ∠CBA = 2 · ∠EX ′B.

But ∠EPB is the angle at the center of Γ2 that is subtended by the chord EB,which is twice any angle on the circumference subtended by EB; consequently, X ′

is a point of Γ2 as well as lying on Γ1. Similarly, X ′ is a point where Γ3 intersectsΓ1, whence X ′ is the point other than E where Γ2 intersects Γ3. That is, X ′ = X,so that AX must be the A-symmedian, as desired.

Editor’s comment. If directed angles are used, then the featured argument doesnot require the given triangle to be acute: the result holds for an arbitrary ∆ABC.

4382. Proposed by Borislav Mirchev and Leonard Giugiuc.

Let ABCD be an orthogonal cyclic quadrilateral with AC ⊥ BD. Let O and Rbe the circumcenter and the circumradius of ABCD respectively and let P be theintersection of AC and BD. Denote by r1, r2, r3 and r4 the inradii of the minorcircular sectors PAB,PBC,PCD and PDA respectively. Prove that

r1 + r2 + r3 + r4 + 8R = (R2 −OP 2)

(1

r1+

1

r2+

1

r3+

1

r4

).


Solutions /357

We received 7 submissions of which 6 were correct and complete. We feature twoof them.


The proof is by coordinates. Set P to be the origin, and let O = (x, y). Assumethe figure is oriented such that B and C lie on the positive y and x-axis, as in theaccompanying figure. Let the center of the incircle of sector PBC be O2.

Note that O2 = (r2, r2), and OO2 = R− r2. Thus, by the distance formula√(x− r2)2 + (y − r2)2 = R− r2 ⇐⇒ x2 + y2 − 2r2(x+ y) + r22 = R2 − 2Rr2

⇐⇒ 2Rr2 − 2r2(x+ y) + r22 = R2 −OP 2,

where the last step follows from OP 2 = x2 + y2. Therefore, we have

R2 −OP 2

r2= 2R+ r2 − 2(x+ y).

Similarly, we may compute

R2 −OP 2

r1= 2R+ r1 + 2(x− y),

R2 −OP 2

r3= 2R+ r3 + 2(y − x)

R2 −OP 2

r4= 2R+ r4 + 2(x+ y).

Upon adding, we arrive at

(R2 −OP 2)

(1

r1+

1

r2+

1

r3+

1

r4

)= r1 + r2 + r3 + r4 + 8R,

which is precisely the desired result.

Solution 2, by Ioannis D. Sfikas.

Let Oi be the center of the incircle whose radius is ri, i = 1, . . . , 4. Assuminglabelling as in the figure, we have

O1O = R− r1, O3O = R− r3, O1P =√

2r1, and O3P =√

2r3;

furthermore, P lies on the line segment O1O3. By Stewart’s theorem applied tothe cevian OP of ∆OO1O3 we have

OP 2 = (R− r3)2r1

r1 + r3+ (R− r1)2

r3r1 + r3

− 2r1r3,

so thatR2 −OP 2

r1r3= 1 +

4R

r1 + r3,


358/ Solutions

and, finally, (R2 −OP 2

)( 1

r1+

1

r3

)= r1 + r3 + 4R.

Similarly, (R2 −OP 2

)( 1

r2+

1

r4

)= r2 + r4 + 4R.

Adding these last two equations gives us the desired result, namely

(R2 −OP 2)

(1

r1+

1

r2+

1

r3+

1

r4

)= r1 + r2 + r3 + r4 + 8R.

Editor’s comments. It is interesting to compare our problem 4382 with Problem1.4.6 on pages 9 and 85 of Japanese Temple Geometry Problems: San Gaku (editedby Hidetosi Fukagawa and Dan Pedoe and published in 1989 by the Charles Bab-bage Research Centre). The Japanese problem applies to essentially the samefigure, except they use a small circle inside the region bounded by arcs of thegiven circles and externally tangent to those four given circles (rather than ourlarge circumcircle that encloses the four given circles and is internally tangent tothem). This requires replacing our radius R by −R. They find a relation amongthe four radii that involves neither R nor OP , namely

r1r3(r2 +r4)2 + r2r4(r1 +r3)2 = (r1r3−r2r4)2 + (r1 +r3)(r2 +r4)(r1r3 +r2r4).

4383. Proposed by Michel Bataille.

Evaluate the integral

I =

∫ 1

0

(lnx) ·√

x

1− x dx.

We received 14 submissions, all correct. We present the solution by Kee-Wai Lau.

We show that the given integral equals

I =(1− 2 ln 2)π

2.

By substitution x = sin2 θ and using half angle formula sin2 θ =1− cos (2θ)

2, we

obtain

I = 4

∫ π/2

0

ln (sin θ) sin2 θ dθ

= 2

(∫ π/2

0

ln (sin θ) dθ −∫ π/2

0

ln (sin θ) cos (2θ) dθ

). (1)


Solutions /359

The first integral in (1) is well know:∫ π/2

0

ln (sin θ) dθ =1

2

∫ π

0

ln (sin θ) dθ = −π ln 2

2

(see, for example, p.246 of G. Boros and V. Moll “Irresistible Integrals”, CambridgeUniversity Press, 2004.)

Since limx→0+ x lnx = 0, using a substitution for the second integral of (1), we get∫ π/2

0

ln (sin θ) cos (2θ) dθ =1

2

∫ π/2

0

ln (sin θ) d(sin (2θ)) = −∫ π/2

0

cos2 θ dθ = −π4.

Combining the integrals, we get that I =(1− 2 ln 2)π

2as claimed.


Let n be an integer with n ≥ 2. Find all real numbers x such that∑0≤i<j≤n−1

⌊x+

i

n

⌋·⌊x+

j

n

⌋= 0.

We received 4 correct solutions. We present the composite solution.

For any real number x and integers i, j with 0 ≤ i < j ≤ n− 1,

bxc ≤⌊x+

i

n

⌋≤⌊x+

j

n

⌋< bx+ 1c = bxc+ 1.

Therefore, each term of the sum assumes exactly one of the values a2, a(a + 1)and (a + 1)2 for a = bxc, and so is non-negative. Therefore the sum vanishes ifand only if each term bx+ i/nc · bx+ j/nc vanishes.

Suppose that −1/n ≤ x < 2/n. Then

0 = bx+ 1/nc = bx+ (n− 2)/nc

so that each summand has at least one factor equal to 0 and the equation issatisfied. On the other hand, when x < −1/n, then

bxc⌊x+

1

n

⌋= (−1)2 > 0

and when x ≥ 2/n, then⌊x+

n− 2

n

⌋·⌊x+

n− 1

n

⌋= 1 > 0.

Therefore the equation is satisfied if and only if −1/n ≤ x < 2/n.


360/ Solutions

4385. Proposed by Miguel Ochoa Sanchez and Leonard Giugiuc.

Let ABC be a triangle with circumcircle ω and AB < AC. The tangent at A toω intersects the line BC at P . The internal bisector of ∠APB intersects the sidesAB and AC at E and F , respectively. Show that

PE

PF=

√EB

FC.

We received 14 correct submissions. We present the solution by K. V. Sudharshan.

Let AD be the internal angle bisector of ∠BAC, with D ∈ BC. We can see that

∠PAD = ∠PAB + ∠BAD = ∠ACB + ∠CAD = ∠PDA.

Thus PA = PD. This implies that EF is the perpendicular bisector of AD.

Since AD bisects ∠EAF , we see that EAFD is a rhombus, and so AE = AF .

Now note that 4PEA ∼ 4PFC, so

PE

PF=AE

FC.

Also, 4PEB ∼ 4PFA and so

PE

PF=BE

AF.

Combining with AE = AF and the previous result, we have AE2 = BE · CF .Thus,

PE

PF=AE

FC=

√BE

CF.

4386. Proposed by Thanos Kalogerakis.

Let ABCD be a cyclic quadrilateral with AD > BC, where X = AB ∩ CD andY = BC ∩AD. The bisectors of angles X and Y intersect BC and CD at P andS, respectively. Finally, let Q and T be points on the sides AD and AB such thatPQ ⊥ AD and ST ⊥ AB. Prove that ABCD is bicentric if and only if PQ = ST .

We received four submissions, but one was withdrawn. We feature the proposer’ssolution, modified by the editor. The proposer clearly intended for the given quadri-lateral to have no parallel sides (rather than demanding inequalities among theedges).

Because a cyclic quadrilateral ABCD has an inscribed circle if and only if thesums of opposite sides are equal, we are to prove that AD + BC = AB + CD ifand only if PQ = ST . We first show that the area of ABCD is 1

2PQ(AD+BC).

From ABCD cyclic we have

∠BAD = ∠BCX. (1)


Solutions /361

Denoting by B′C ′ the reflection of the segment BC in the mirror XP , the trianglesPBC ′ and PB′C are congruent (since P is fixed by the reflection) and, therefore,have the same area. It follows that the quadrangles ABCD and AC ′B′D have thesame area. Furthermore we have

∠BCX = ∠B′C ′X and BC = B′C ′. (2)

From (1) and (2) we have ∠BAD = ∠BCX = ∠B′C ′X, so that the lines ADand B′C ′ are parallel and, therefore, AC ′B′D is a trapezoid with altitude PQ andarea

1

2PQ(AD +B′C ′).

Because this is also the area of our given quadrilateral and (by (2)) BC = B′C ′,the area of ABCD is 1

2PQ(AD +BC), as claimed.

Similarly, using the same argument with X, XP , PQ, AD, BC replaced by Y , Y S,ST , AB, CD, we deduce that the area of ABCD is also given by 1

2ST (AB+CD).That is,

PQ(AD +BC) = ST (AB + CD),

from which we conclude that AD + BC = AB + CD if and only if PQ = ST , asdesired.

Editor’s Comment. If ABCD has an incircle, its diameter has length PQ = ST .This follows from the observations that the bisector of ∠X must pass throughthe incenter, which implies that the reflection in this line fixes the incircle andtherefore takes its tangent BC to another tangent B′C ′; the length of the commonperpendicular PQ is the distance between parallel tangents AD and B′C ′ which,of course, is the diameter of the incircle.

4387. Proposed by Nguyen Viet Hung.

Let

an =

n∑k=1

k

√1 +

k2

(k + 1)!, n = 1, 2, 3, . . .

Determine banc and evaluate limn→∞

ann.

We received 7 submissions, of which 6 were correct and complete. We present thesolution by Sushanth Sathish Kumar.

We claim that banc = n. Clearly, an ≥ n, since each term in the sum is at least1. By the generalized Bernoulli inequality, (1 + x)1/k ≤ 1 + x

k for any k ∈ N and


362/ Solutions

x ≥ −1, so in particular we have

an =

n∑k=1

(1 +

k2

(k + 1)!

) 1k

≤n∑k=1

(1 +

k

(k + 1)!

)

=

n∑k=1

1 +

n∑k=1

(1

k!− 1

(k + 1)!

)= n+

(1− 1

(n+ 1)!

).

Thus, n ≤ an < n+ 1, which implies banc = n. By the squeeze theorem,

1 = limn→∞

n

n≤ limn→∞

ann≤ limn→∞

n+ 1

n= 1;

that is, limn→∞

ann = 1.

4388. Proposed by Marian Cucoanes and Leonard Giugiuc.

For positive real numbers a, b and c, prove

8abc(a2 + 2ac+ bc)(b2 + 2ab+ ac)(c2 + 2bc+ ab) ≤ [(a+ b)(b+ c)(c+ a)]3.

We received 6 submissions, including the one from the proposers. One of the givensolutions used Maple outputs. We present the proof by Daniel Vacaru.

The given inequality is equivalent to

(a2 + 2ac+ bc)(b2 + 2ba+ ca)(c2 + 2cb+ ab)

(a+ b)2(b+ c)2(c+ a)2≤ (a+ b)(b+ c)(c+ a)

8abc(1)

Let E denote the LHS of (1). Then by direct computations we have

E =( a

a+ b+

c

c+ a

)( b

b+ c+

a

a+ b

)( c

c+ a+

b

b+ c

). (2)

Since a+ b ≥ 2√ab, b+ c ≥ 2 +

√bc, c+ a ≥ 2

√ca, we get from (2) that

E ≤ 1

8

(√a

b+

√c

a

)(√b

c+

√a

b

)( c

c+ a+

b

b+ c

)=

1

8abc(a+

√bc)(b+

√ca)(c+

√ab)

=1

8abc

(2abc+ ab

√ab+ bc

√bc+ ca

√ca+ a2

√bc+ b2

√ca+ c2

√ab). (3)


Solutions /363

Using AM-GM Inequality again we then obtain from (3) that

E ≤ 1

8abc

(2abc+ ab

(a+ b

2

)+ bc

(b+ c

2

)+ ca

(c+ a

2

)+a2

(b+ c

2

)+ b2

(c+ a

2

)+ c2

(a+ b

2

))=

1

8abc

(2abc+ a2b+ b2c+ c2a+ ab2 + bc2 + ca2

)=

1

8abc(a+ b)(b+ c)(c+ a),

so (1) holds and the proof is complete.

4389. Proposed by Daniel Sitaru.

Consider the real numbers a, b, c and d. Prove that

a(c+ d)− b(c− d) ≤√

2(a2 + b2)(c2 + d2).

We received 21 solutions, all correct, and will feature the solution by MichelBataille.

The inequality certainly holds if a(c+d)− b(c−d) < 0 and otherwise is equivalentto

(ac+ ad− bc+ bd)2 ≤ 2(a2 + b2)(c2 + d2).

Now, a simple calculation shows that

2(a2 + b2)(c2 + d2)− (ac+ ad− bc+ bd)2 = (ac+ bd− ad+ bc)2 ≥ 0

so we are done.

4390. Proposed by Marius Dragan and Neculai Stanciu.

Let x, y and z be positive real numbers with x + y + z = m. Find the minimumvalue of the expression

1

1 + x2+

1

1 + y2+

1

1 + z2.

We received 5 submissions of which 2 were correct and complete. We present thesolution by the proposers, with minor edits.

We use the following result.

Let s > 0 and let F (x1, x2, . . . , xn) be a symmetrical continuous function on thecompact set in Rn

S = {(x1, x2, . . . , xn) : x1 + x2 + · · ·+ xn = s, x1 ≥ 0, . . . , xn ≥ 0}.


364/ Solutions

If

F (x1, x2, . . . , xn) ≥ min{F

(x1 + x2

2,x1 + x2

2, x3, . . . , xn

),

F (0, x1 + x2, x3, . . . , xn)}

(1)

for all (x1, . . . , xn) ∈ S with x1 > x2 > 0, then

F (x1, . . . , xn) ≥ min1≤k≤n

F( sk, . . . ,

s

k, 0, . . . , 0

),

for all (x1, . . . , xn) ∈ S.

A proof of this can be found in Algebraic inequalities, Old and New Methods, V.Cartoaje, Gil Publishing house, 2006, 267-269.

Let F (x, y, z) = 11+x2 + 1

1+y2 + 11+z2 , and let x, y, z be positive reals such that

x+ y + z = m and x > y > 0.

We verify (1) by proving that

if F (x, y, z) < F(x+y2 , x+y2 , z

), then F (x, y, z) ≥ F (0, x+ y, z).

Put t = x+y2 , and p = xy. Rearranging the inequality F (x, y, z) < F

(x+y2 , x+y2 , z

),

gives(t2 − p)(4t2 + 2p− 2)

(1 + x2)(1 + y2)(1 + t2)< 0,

and since t2−p > 0, we have 4t2 < 2−2p. It follows that 4t2p < 2p−2p2 ≤ 2−2p,and so 2−4t2p−2p > 0. The inequality F (x, y, z) ≥ F (0, x+y, z) can be rearrangedto obtain the equivalent inequality

xy(2− 4t2p− 2p)

(1 + x2)(1 + y2)(1 + t2)≥ 0,

which follows from the above. Thus we can apply the cited result, which gives

F (x, y, z) ≥ min{F (m, 0, 0), F

(m2,m

2, 0), F(m

3,m

3,m

3

)}= min

{2m2 + 3

m2 + 1,m2 + 12

m2 + 4,

27

m2 + 9

}

=

2m2 + 3

m2 + 1, m ∈ (0,

√2]

m2 + 12

m2 + 4, m ∈ (

√2,√

6]

27

m2 + 9, m ∈ (

√6,∞).


Solutions /365

4391. Proposed by Leonard Giugiuc and Oai Thanh Dao.

Let ABC be an equilateral triangle and let W be a point inside ABC. A linel1 through W intersects the segments BC and AB in D and P , respectively.Similarly, a line l2 through W intersects AC and BC in E and M , and a line l3through W intersects AB and AC in F and N . If

∠DWE = ∠EWF = ∠FWD = 120◦,

show that the triangles DEF and MNP are similar.

We received 6 solutions. We present the solution by C. R. Pranesachar.

It is easy to see that each of the six angles made by l1, l2 and l3 at W is 60◦. ThusMWNC, DWEC and EWFA each have opposite angles which add up to 180◦,and are hence cyclic quadrilaterals. So

∠WNM = ∠WCM = ∠WCD = ∠WED, and

∠WNP = ∠WAP = ∠WAF = ∠WEF.

Adding, we get ∠MNP = ∠DEF . Similarly, ∠NPM = ∠EFD and ∠PMN =∠FDE. Thus triangles DEF and MNP are similar.

Editor’s Comments. As noted by J. Chris Fisher, this problem is a special con-sequence of Miquel’s theorem. Namely, in any triangle ABC, if we choose pointsD on BC, E on AC and F on AB, the circumcircles of the three triangles DCE,EAF and FBD intersect at one point, called the Miquel point for DEF . If MNPis constructed analogously and has the same Miquel point as DEF then MNP issimilar to DEF . For more details, see Roger A. Johnson’s “Advanced EuclideanGeometry”, paragraphs 183-188, and problem 1992: 176, 1993: 152-153 proposedby J. Chris Fisher, Dan Pedoe and Robert E. Jamison.


366/ Solutions

4392. Proposed by Leonard Giugiuc and Kadir Altintas.

Let M be an interior point of a triangle ABC with sides BC = a, CA = b andAB = c. If MA = x,MB = y and MC = z, then prove that if√

(a+ y − z)(a− y + z) +√

(b+ z − x)(b− z + x) +√

(c+ x− y)(c− x+ y)

=√

3(x+ y + z),

then ABC is equilateral.

We received 5 solutions, 4 of which were correct. We present the solution bySushanth Sathish Kumar.

By Cauchy-Schwarz, we may estimate

a+ b+ c =√

[(a+ y − z) + (b+ z − x) + (c+ x− y)]

·√

[(a− y + z) + (b− z + x) + (c− x+ y)]

≥√

(a+ y − z)(a− y + z) +√

(b+ z − x)(b− z + x)

+√

(c+ x− y)(c− x+ y)

=√

3(x+ y + z)

But on the other hand, the estimate√(a+ y − z)(a− y + z) =

√a2 − (y − z)2 ≥ a,

gives that √3(x+ y + z) ≥ a+ b+ c.

Thus, it follows thata+ b+ c =

√3(x+ y + z).

Additionally, we have x = y = z, since otherwise equality cannot occur in thesecond estimate.

Since x = y = z, M must be the circumcenter of ABC. Hence, x = y = z = R,where R is the circumradius of the triangle. So the equation

a+ b+ c =√

3(x+ y + z)

reduces to

sinA+ sinB + sinC =a

2R+

b

2R+

c

2R=

3√

3

2.

But Jensen’s inequality implies that

sinA+ sinB + sinC ≤ 3 sin

(A+B + C

3

)= 3√

3/2,

with equality holding if and only if A = B = C = π/3. Consequently, equalitymust occur in the above equation, which gives A = B = C = π/3, or that ABC isequilateral.


Solutions /367

4393. Proposed by Ruben Dario Auqui and Leonard Giugiuc.

Let ABCD be a square. Find the locus of points P inside ABCD such thatcot∠CPD + cot∠CDP = 2.

We received 13 solutions, all correct, We feature 4 of them here.

Solution 1, by Paul Bracken and Sushanth Sathish Kumar (done independently).

Assign coordinates A ∼ (0, 0), B ∼ (0, 1), C ∼ (1, 1), D ∼ (1, 0), P ∼ (x, y), andlet γ, δ and θ be the respective angles at C, D and P in triangle CDP . Then

cot γ =1− y1− x, cot δ =

y

1− x,

and

cot θ =1− cot γ cot δ

cot γ + cot δ=

(1− x)2 + (y2 − y)

1− x .

Therefore

2(1− x) = (1− x)(cot θ + cot δ) = (1− 2x+ x2) + y2,

whence x2+y2 = 1. The locus of P is a quarter-circle centred at A passing throughthe vertices B and D.

Solution 2, by Michel Bataille.

Let c, d, p, r be the respective lengths of DP,PC,CD,PA; let h and k be therespective distances from P to CD and AD; let γ, δ, θ be the respective angles atC, D, P in triangle CPD, and let S be the area of triangle CPD.

Noting that2S = pd sin γ = cd sin θ = cp sin δ = ph,

we have that

2 =cos θ

sin θ+

cos δ

sin δ=

sin(θ + δ)

sin θ sin δ=

sin γ

sin θ sin δ=

c2(pd sin γ)

(cd sin θ)(cp sin δ)=

c2

2S,

whence c2 = 4S = 2ph.


368/ Solutions

Thenr2 = k2 + (p− h)2 = (c2 − h2) + (p2 − 2ph+ h2) = p2.

Therefore, for every position of P , its distance from A is equal to p, so that thelocus of P is a quarter-circle with centre A passing through B and D.

Solution 3, by C.R. Pranesachar.

From the Cosine Law, we obtain for any triangle ABC,

cotA =b2 + c2 − a2

2bc sinA=b2 + c2 − a2

4[ABC],

where [ABC] is the area of the triangle. Applying this to triangle CPD and usingthe notation of the previous solution, we have that

2 = cot θ + cot δ =(c2 + d2 − p2) + (c2 + p2 − d2)

4[CPD]=

2c2

2ph.

Thus c2 = 2ph and we can complete the argument as in Solution 2.

Solution 4, by Vaclav Konecny.

Adopt the notation of Solution 2. Let Q and R be the respective feet of theperpendiculars from A and C to the line DP . Then ∠QAD = δ and so the lengthof QD is equal to p sin δ.

The length p of PD is equal to the sum of the lengths of PR and RD when θ ≤ 90◦

and to the difference of these lengths when θ > 90◦. The length of RD is equal top cos δ. The length of PR is equal to p sin δ cot θ when θ ≤ 90◦ and −p sin δ cot θwhen θ > 90◦. In any case, we find that the length of PD is equal to

p cos δ + p sin δ cot θ = p cos δ + p sin δ(2− cot δ) = 2p sin δ,

i.e. twice the length of QD. It follows that the triangle PAD is isosceles and sothe lengths of AP and AD are both equal to p.

Therefore P lies on the quarter-circle with centre A through B and C.

4394. Proposed by Mihaela Berindeanu.

Let ABC be an acute triangle and M ∈ BC, BM ≡ MC, E ∈ AB, F ∈ AC,]BEM ≡ ]CFM = 90◦. The two tangents at the points E and F to thecircumcircle of 4MEF intersect at the point X. If XM ∩ EF = {Y }, show thatY B = Y C.

We received 4 submissions, all correct, and feature two of them.

Solution 1, by Shuborno Das.

To show Y B = Y C, it suffices to show that YM ⊥ BC. But since X,M , andY are collinear, we just need to prove that XM ⊥ BC. As X is the intersection


Solutions /369

of the tangents to the circle MEF at E and F , MX is the M -symmedian in∆MEF . Let’s consider a mapping Ψ that is the product of the inversion in thecircle centered at M with radius

√ME ·MF followed by the reflection in the

bisector of ∠EMF . Note that E and F are interchanged by Ψ. As in the figure,we shall use a prime to denote the image of a point under Ψ.

Because ∠MEB = 90◦, we have ∠MB′E′ = ∠MB′F = 90◦, and similarly,∠MC ′E = 90◦. Since MB = MC and M ∈ BC, it follows that MB′ = MC ′ andM ∈ B′C ′. Moreover, because MX is the M -symmedian in ∆MEF , MX ′ is theM -median in ∆MEF . Because lines through M are sent by Ψ to their reflectionin the bisector of ∠EMF , our problem is reduced to showing that MX ′ ⊥ B′C ′.Let MX ′ meet EF at N ; then NE = NF . But MB′ = MC ′ and EC ′||FB′(because both lines are perpendicular to B′C ′), while MN is the line that joinsthe midpoints of B′C ′ and FE, whence NM ⊥ B′C ′, and we are done.


As in Solution 1, we just need to prove that XM ⊥ BC.

Define U = FM ∩ AE and V = EM ∩ AF . Let X ′ be the midpoint of UV . Weclaim that X = X ′: note that ∠UEV = ∠BEM = 90◦ = ∠MFC = ∠UFV , soX ′ is the circumcenter of cyclic quadrilateral UEFV . As such, triangles X ′UE,X ′EF , and X ′FV are all isosceles with vertex X ′.


370/ Solutions

Abbreviating ∠AUV = U , ∠AV U = V and angle chasing yields

∠EX ′F = 180◦ − ∠UX ′E − ∠FX ′V

= 180◦ − (180◦ − 2U)− (180◦ − 2V )

= 2U + 2V − 180◦.

Thus, from triangle X ′EF

∠X ′EF = ∠X ′FE = 180◦ − U − V = ∠UAV = ∠EAF,

which implies that X ′E and X ′F are tangent to (AEMF ). Moreover, this impliesX is the circumcenter of (UEFV ). Let rays MB, and MC meet (UEFV ) atpoints P , and Q, respectively. By the converse to the butterfly theorem, M isthe midpoint of the chord PQ, and (recalling that X is the center of the circle(UEFV )) we see this implies that XM ⊥ PQ. Consequently, XM ⊥ BC, asdesired.

We end with a projective proof of the converse of the butterfly theorem. Perspec-tivities from points E and F of the circle (UPEFUV ) give

(P,M ;B,Q)E= (P, V ;U,Q)

F= (P,C;M,Q),

where (P,M ;B,Q), is the cross-ratio. Hence (using the hypothesis BM = MC),

BP

BM· QMQP

=MP

MC· QCQP

=⇒ (BP )(QM) = (MP )(QC).

Taking into account that BP = MP −MB, we get

(MP−MB)(QM) = (MP )(QC) =⇒ MB(QM) = MP (QM−QC) = MP (MC),

which gives MP = MQ; thus, M is the midpoint of chord PQ and we are done.


Solutions /371

Editor’s comments. This editor was unable to find an explicit statement of thebutterfly theorem’s converse. Many of the vast variety of published proofs ofthe theorem are reversible, and thus the converse has been tacitly, yet firmly,established. See, for example the first of the editor’s proofs in the second volumeof Crux (when the journal was called Eureka) [1976: 2-3], or about half of themany proofs presented by Leon Bankoff in “The Metamorphosis of the ButterflyProblem” [Mathematics Magazine, 60:4 (Oct. 1987) 195-210], including a versionof the argument above in Solution 2.


Let ABCD be a tetrahedron and let A1, B1, C1, A2, B2, C2 be the midpoints ofBC,CA,AB,DA,DB,DC, respectively. Prove that

(−−→DA · −−→BC)A1A

22 + (

−−→DB · −→CA)B1B

22 + (

−−→DC · −−→AB)C1C

22 = 0,

where−→X · −→Y denotes the dot product of the vectors

−→X and

−→Y .

We received 7 solutions, all of which were correct. We present the solution byOliver Geupel.

Consider location vectors relative to the origin at point D. We have

2−→A1 =

−→B +−→C , 2

−→B1 =

−→C +−→A, 2

−→C1 =

−→A +−→B, 2

−→A2 =

−→A, 2

−→B2 =

−→B, 2

−→C2 =

−→C .

We use the alternative notation⟨−→X,−→Y⟩

for the inner product of vectors−→X and

−→Y for reasons of readability. Let

a =⟨−→B,−→C⟩, b =

⟨−→C ,−→A⟩, c =

⟨−→A,−→B⟩.

It follows that

4⟨−−→DA,

−−→BC

⟩A1A

22 =

⟨−→A,−→C −−→B

⟩(−→A −−→B −−→C

)2= (b− c)

(−→A 2 +

−→B 2 +

−→C 2 + 2(a− b− c)

)= (b− c)

(−→A 2 +

−→B 2 +

−→C 2)

+ 2(c2 − b2 + ab− ac

).

With similar identities for the other two terms, we obtain

4(⟨−−→DA,

−−→BC

⟩A1A

22 +

⟨−−→DB,

−→CA⟩B1B

22 +

⟨−−→DC,

−−→AB⟩C1C

22

)= ((b− c) + (c− a) + (a− b))

(−→A 2 +

−→B 2 +

−→C 2)

+ 2(c2 − b2 + ab− ca

)+ 2

(a2 − c2 + bc− ab

)+ 2

(b2 − c2 + ca− bc

)= 0,

which proves the desired identity.


372/ Solutions

4396. Proposed by David Lowry-Duda.

Show that there is a bijection f : N 7→ N such that the series∑∞n=1

1

n+ f(n)converges or show that no such bijection exists.

We received 7 submissions, of which 6 were correct and complete. We present thesolution by Oliver Geupel.

Such a bijection exists. For n ∈ N, let an = n2 be the sequence of all perfectsquares in ascending order and let

b1 = 2, b2 = 3, b3 = 5, b4 = 6, b5 = 7, b6 = 8, b7 = 10, . . .

be the sequence of all positive non-squares in ascending order. Define f : N→ N

by f(an) = bn and f(bn) = an for all n ∈ N. This defines a bijection because Nis the disjoint union of {an : n ∈ N} and {bn : n ∈ N}. For every k ∈ N each pair{ak, bk} appears as {n, f(n)} for exactly two distinct values of n ∈ N. We get

∞∑n=1

1

n+ f(n)= 2

∞∑n=1

1

an + bn< 2

∞∑n=1

1

an= 2

∞∑n=1

1

n2=π2

3.

Consequently, the series is convergent.

4397. Proposed by George Stoica.

Let n ∈ N and k ∈ {0, 1, . . . , 2n}. Show that there exists k′ ∈ {0, 1, . . . , 2n+1}such that ∣∣∣∣sin k′π

2n+2− k

2n

∣∣∣∣ ≤ 1

2n.

We received 2 solutions, both of which were correct. We present the solution byOmran Kouba.

Letam = sin

( πm2n+2

)for m ∈ {0, 1, . . . , 2n+1}.

Clearly,0 = a0 < a1 < · · · < am < am+1 < · · · < a2n = 1.

Further, for 0 ≤ x ≤ y ≤ π/2, we have

0 ≤ sin y − sinx =

∫ y

x

cos t dt ≤∫ y

x

dt ≤ y − x.

So, since π < 4, we have

am+1 − am <π

2n+2<

1

2n, for m = 0, 1, . . . , 2n+1 − 1.

Now, given k ∈ {0, 1, . . . , 2n} we consider

Nk =

{m ∈ {0, 1, . . . , 2n+1} : am ≤

k

2n

}.


Solutions /373

Clearly, Nk 6= ∅ because it contains 0 and it has 2n+1 as an upper bound. So wemay define k′ = maxNk.

• If k′ = 2n+1 then ak′ = 1; this corresponds to the case k = 2n and thedesired inequality holds trivially in this case.

• If k′ < 2m+1 then by definition of k′ we have ak′ ≤ k2−n < ak′+1 so

0 <k

2n− ak′ < ak′+1 − ak′ <

1

2n.

and the desired inequality follows.


Prove that for n ∈ N∗, we have

1

2n− 1+

∫ 1

0

sin2(xn)dx ≥ 2

n(1− cos 1).

We received 10 submissions, including the one from the proposer, all of whichare correct. We present the nearly identical solution by Michel Bataille, LeonardGiugiuc, Digby Smith, and Daniel Vacaru.

Since a2 + b2 ≥ 2ab for all a, b ∈ R, we have

1

2n− 1+

∫ 1

0

sin 2(xn)dx =

∫ 1

0

(x2n−2 + sin2(xn))dx

≥ 2

∫ 1

0

xn−1 sin (xn)dx

= − 2

ncos (xn)

∣∣∣∣10

= − 2

n(cos 1− cos 0)

=2

n(1− cos 1).

4399. Proposed by Lacin Can Atis.

Let ABCDE be a pentagon. Prove that

|AB||EC||ED|+ |BC||EA||ED|+ |CD||EA||EB| ≥ |AD||EB||EC|.

When does equality hold?

We received 7 solutions. We present the one by Michel Bataille.


374/ Solutions

We shall use the notation XY instead of |XY |. We observe that the left-hand sideL of the inequality rewrites as

L = ED · (AB · EC + EA ·BC − EB ·AC)

+ EB · (ED ·AC + EA · CD −AD · EC) +AD · EB · EC.

From Ptolemy’s inequality, we have

AB · EC + EA ·BC − EB ·AC ≥ 0

with equality if and only if A, B, C, E lie, in this order, on a circle and

ED ·AC + EA · CD −AD · EC ≥ 0

with equality if and only if A, C, D, E lie, in this order, on a circle.

It follows that L ≥ AD · EB · EC, the desired inequality, and that equality holdsif and only if ABCDE is a cyclic pentagon.


Prove that in any triangle ABC, the following relationship holds:∑cyc

sin(π3 − A2 ) sin(π3 − B

2 )

cos(C−A2 ) cos(C−B2 )< 1.

We received 6 submissions, including the one from the proposer. As it turned out,the proposed inequality is false, and four of the five solvers give various counterex-amples. We will feature below some of the given solutions and comments.

Counterexample 1, given by Leonard Giugiuc, enhanced by the editor.

Let E denote the LHS of the given inequality. Set A = B ∈ (0, π2 ) so C = π− 2A.Then


2 )

cos(C−A2 ) cos(C−B2 )=

sin2(π3 − A2 )

cos2(π−3A2 ), (1)

sin(π3 − B2 ) sin(π3 − C

2 )

cos(A−B2 ) cos(A−C2 )=

sin(π3 − A2 ) sin(A− π

6 )

cos( 3A−π2 )

=sin(π3 − A

2 ) sin(A− π6 ) cos(π−3A2 )

cos2(π−3A2 ), (2)

and

sin(π3 − C2 ) sin(π3 − A

2 )

cos(B−C2 ) cos(B−A2 )=

sin(A− π6 ) sin(π3 − A

2 )

cos( 3A−π2 )

=sin(A− π

6 ) sin(π3 − A2 ) cos( 3A−π

2 )

cos2(π−3A2 ). (3)


Solutions /375

From (1)+(2)+(3) we then obtain

E =1

cos2(π−3A2 )

(sin2

(π3− A

2

)+ 2 sin

(π3− A

2

)sin(A− π

6

)cos(3A− π

2

)).

Now, let A→ 0+. Then

cos2(π − 3A

2

)→ 0+, sin

(π3− A

2

)→√

3

2, and sin

(A− π

6

)→ −1

2

solimA→0+

E =∞.

Counterexample 2, by Alexandru Daniel Pırvuceanu, with all the details suppliedby the editor.

Let A = 150◦, B = C = 15◦. Then with calculations carried to 4 decimal places,we have:


2 )

cos(C−A2 ) cos(C−B2 )=−(sin 15◦)(sin 52.5◦)

cos(67.5◦) cos 0◦= −0.5366

sin(π3 − B2 ) sin(π3 − C

2 )

cos(A−B2 ) cos(A−C2 )=

sin2(52.5◦)

cos2(67.5◦)= 4.2980

sin(π3 − C2 ) sin(π3 − A

2 )

cos(B−C2 ) cos(B−A2 )=

sin(52.5◦)(− sin 15◦)

cos(67.5◦)= −0.5366

Hence, E = 4.2980− 2(0.5366) = 3.2248 > 1.

Editor’s note. Digby Smith remarked that the given inequality holds if each ofA,B,C > π

6 , and Pranesachar gave examples to show that E < 1, E = 1, andE > 1 are all possible.


SOLUTIONS - Canadian Mathematical Society · 356/ Solutions We received 6 submissions, all of which...

Documents

Transcript of SOLUTIONS - Canadian Mathematical Society · 356/ Solutions We received 6 submissions, all of which...