Volume/tome 50, issue/numéro 2

February/février 2024
Crux Mathematicorum is a problem-solving journal at the secondary and university undergraduate levels,
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Editorial Board

Editor-in-Chief Kseniya Garaschuk University of the Fraser Valley

MathemAttic Editors John Grant McLoughlin University of New Brunswick

Shawn Godin Cairine Wilson Secondary School
Olympiad Corner Editors Alessandro Ventullo University of Milan
Anamaria Savu University of Alberta
Articles Editor Robert Dawson Saint Mary’s University
Associate Editors Edward Barbeau University of Toronto
Chris Fisher University of Regina
Edward Wang Wilfrid Laurier University
Dennis D. A. Epple Toronto, Canada
Magdalena Georgescu Toronto, Canada
Chip Curtis Missouri Southern State University
Philip McCartney Northern Kentucky University
Guest Editors Yagub Aliyev ADA University, Baku, Azerbaijan
Ana Duff Ontario Tech University
Mateusz Buczek Warsaw, Poland
Andrew McEachern York University
Vasile Radu Birchmount Park Collegiate Institute
Chi Hoi Yip University of British Columbia
Matt Olechnowicz Concordia University
Translators Rolland Gaudet Université de Saint-Boniface
Frédéric Morneau-Guérin Université TÉLUQ
Editor-at-Large Bill Sands University of Calgary
 IN THIS ISSUE / DANS CE NUMÉRO

59 MathemAttic: No. 52
59 Problems: MA256–MA260
61 Solutions: MA231–MA235
66 Teaching Problems: No. 25 Ed Barbeau
69 Olympiad Corner: No. 420
69 Problems: OC666–OC670
71 Solutions: OC641–OC645
77 Reading a Math Book: Solutions to No. 2 Yagub Aliyev
82 Problems: 4911–4920
87 Solutions: 4861–4870

Crux Mathematicorum
Founding Editors / Rédacteurs-fondateurs: Léopold Sauvé & Frederick G.B. Maskell
Former Editors / Anciens Rédacteurs: G.W. Sands, R.E. Woodrow, Bruce L.R. Shawyer,
Shawn Godin

Crux Mathematicorum
with Mathematical Mayhem
Former Editors / Anciens Rédacteurs: Bruce L.R. Shawyer, James E. Totten, Václav Linek,
Shawn Godin
 MathemAttic /59

MATHEMATTIC
No. 52
The problems in this section are intended for students at the secondary school level.

Click here to submit solutions, comments and generalizations to any

problem in this section.

To facilitate their consideration, solutions should be received by April 15, 2024.

MA256. There exist positive integers whose value is quadrupled by moving

the rightmost decimal digit into the leftmost position. Find the smallest such
number.

MA257. A square of area 1 is divided into three rectangles which are geomet-
rically similar (i.e., they have the same ratio of long to short sides) but no two of
which are congruent. Let A, B and C be the areas of the rectangles, ordered from
largest to smallest. Prove that (AC)2 = B 5 .

MA258. The three following circles are tangent to each other: the first has
centre (0, 0) and radius 4, the second has centre (3, 0) and radius 1, and the third
has centre (−1, 0) and radius 3. Find the radius of a fourth circle tangent to each
of these 3 circles.

MA259. Consider the equation 7a + 12b = c where a, b and c are nonnegative

integers. For many values of c, it is possible to find one or more pairs (a, b)
satisfying the equation. Given c = 26, for example, (a, b) = (2, 1) is the only
solution.
a) If c = 365, find all possible solutions (a, b), where a and b are nonnegative
integers.
b) There are some values of c for which no solutions exist. For example, there
is no pair (a, b) such that 7a + 12b = 20, so c = 20 is one such case. Find the
largest integer value of c for which there are no nonnegative integer solutions.

MA260. The expression n! denotes the product 1 · 2 · 3 · · · n and is read as

“n factorial”. For example, 5! = 1 · 2 · 3 · 4 · 5 = 120.
a) The product (2!)(3!)(4!)(5!)(6!)(7!)(8!)(9!)(10!)(11!)(12!) can be written in
the form M 2 N !, where M , N are positive integers. Find a suitable value of
N and justify your answer.
b) Prove that, for every n ≥ 1, (2!)(3!)(4!) · · · ((4n)!) can be written as the
product of a square and a factorial.

Copyright © Canadian Mathematical Society, 2024

60/ MathemAttic

Les problèmes proposés dans cette section sont appropriés aux étudiants de l’école sec-
ondaire.

Cliquez ici afin de soumettre vos solutions, commentaires ou

généralisations aux problèmes proposés dans cette section.

Pour faciliter l’examen des solutions, nous demandons aux lecteurs de les faire parvenir
au plus tard le 15 avril 2024.

MA256. Il existe des entiers positifs dont la valeur est quadruplée lorsque l’on
déplace le chiffre le plus à droite vers la position la plus à gauche. Trouver le plus
petit nombre ayant cette propriété.

MA257. Un carré d’aire 1 est divisé en trois rectangles qui sont géométriquement
similaires (i.e., les rapports entre les longueurs des côtés longs sur les longueurs
des côtés courts sont les mêmes) mais aucun de ces rectangles est congru à un
autre. Nommer A, B et C les aires des rectangles ordonnés du plus grand au plus
petit. Prouver que (AC)2 = B 5 .

MA258. Trois cercles sont tangents les uns aux autres; le premier a son
centre à (0, 0) et son rayon est 4; le second a son centre à (3, 0) et son rayon est
1; le troisime a son centre à (−1, 0) et son rayon est 3. Déterminer le rayon d’un
quatrième cercle, tangent à chacun de ces 3 cercles.

MA259. Considérer l’équation 7a + 12b = c, où a, b et c sont des entiers

nonnégatifs. Étant donné c, il y a parfois une solution ou plus, (a, b), à l’équation.
Par exemple, si c = 26, alors (2, 1) est la seule solution (a, b).
a) Si c = 365, déterminer toutes les solutions possibles (a, b), où a et b sont des
entiers non négatifs.
b) Il existe certaines valeurs de c pour lesquelles il n’y a aucune solution (a, b)
à l’équation précédente. Par exemple, si c = 20, l’équation 7a + 12b = 20 n’a
aucune solution (a, b). Déterminer la plus grande valeur de c pour laquelle
il n’existe aucune solution (a, b) où a et b sont des entiers non négatifs.
MA260. L’expression n! représente le produit 1 · 2 · 3 · · · n et se lit “factorielle
n”. Par exemple, 5! = 1 · 2 · 3 · 4 · 5 = 120.
a) Le produit (2!)(3!)(4!)(5!)(6!)(7!)(8!)(9!)(10!)(11!)(12!) peut s’écrire sous la
forme M 2 N !, où M et N sont des entiers strictement positifs. Déterminer
une valeur appropriée de N et justifier votre réponse.
b) Montrer que pour tout n ≥ 1, (2!)(3!)(4!) · · · ((4n)!) peut s’écrire comme le
produit d’un carré et d’une factorielle.

Crux Mathematicorum, Vol. 50(2), February 2024

 MathemAttic /61

MATHEMATTIC
SOLUTIONS
Statements of the problems in this section originally appear in 2023: 49(7), p. 340–341.

MA231. A grocery store clerk wants to make a large triangular pyramid of

oranges. The bottom level is an equilateral triangle made up 1275 oranges. Each
orange above the first level rests in a pocket formed by three oranges below. The
stack is completed at the final level with a single orange. How many oranges are
in the stack?
Originally question 9 from the 35th University of Alabama High School Mathemat-
ics Tournament: Team Competition, 2016.
We received 7 submissions, all correct and complete. We present the solution by
Teodor Constantin.
What is the smallest orange pyramid that one can stack with more than one
orange? If one takes 3 oranges to form the equilateral triangular base, and adds
one more orange on top, a 4 orange pyramid is formed! In this case, n = 2 oranges
were used for the side of the bottom triangle.
How about when n = 3? Since n = 3 is the number of oranges that can fit on
the side of the equilateral triangle of the pyramid bottom, one can only fit 3 rows
of oranges in this triangle, where the first row has 3 oranges, the second one has
2 oranges, and the third one has only 1 orange. Note that the total number or
oranges that can be arranged in the bottom of the pyramid is 3+2+1 or 3·4/2, i.e.,
6 oranges in all. As for the total number of oranges in the pyramid, it is important
to note that 3 layers of oranges can be arranged in the following distribution:
3rd layer (bottom) 2nd layer 1st layer (top)
3+2+1 2+1 1

Similarly, when n = 4, the bottom of the pyramid will contain 4 + 3 + 2 + 1 or

4 · 5/2, i.e., 10 oranges in all. The pyramid will have 4 layers with the oranges
being stacked as shown in the following table:
4th layer (bottom) 3rd layer 2nd layer 1st layer (top)
4+3+2+1 3+2+1 2+1 1

It is important to notice that regardless if n is odd or even, the pyramid will always
contain n layers of oranges, with a decreasing number of oranges from bottom to
top with only one orange on the top layer.
To generalize, for an arbitrary number n of oranges that can be arranged on the
side of the triangle, the total number of oranges needed to cover the bottom layer
will be n + (n − 1) + · · · + 2 + 1. This sum is equal to n(n+1)
2 . Given that the

Copyright © Canadian Mathematical Society, 2024

62/ MathemAttic

number of oranges on the bottom level is 1275, the following equality is obtained:
n(n+1)
2 = 1275. This quadratic equation can be written as (n + 51)(n − 50) = 0,
which has only one positive solution, i.e. n = 50. The second solution, n = −51,
is not a viable solution since it’s a negative integer.
To find out the total number of oranges in the pyramid, the number of oranges in
all layers has to be computed. The distribution of oranges is shown below:
50th layer (bottom) 49th layer ··· 1st layer (top)
50 + 49 + · · · +1 49 + 48 + · · · +1 ··· 1

Let S be the total number of oranges in the pyramid. The sum can be written as:

S = 1 · 50 + 2 · 49 + 3 · 48 + · · · + 49 · 2 + 50 · 1,

which is equivalent to

S = 1 · (51 − 1) + 2 · (51 − 2) + 3 · (51 − 3) + · · · + 49 · (51 − 2) + 50 · (51 − 50),

which forces number 51 to appear in all terms of the sum, as well as a distribution
of perfect squares: 1, 22 , · · · , 502 . Therefore the sum becomes:

S = 51 · (1 + 2 + · · · + 50) − (12 + 22 + · · · + 502 ).

Let S1 = 1 + 2 + · · · + 50 and S2 = 12 + 22 + · · · + 502 . Then S = 51 · S1 − S2 .

The values for S1 and S2 can be found by using standard summation formulas, so
that S becomes:
50 · 51 50 · 51 · 101 50 · 51 · 52
S = 51 − = .
2 6 6
Therefore, the total number of oranges in the triangular pyramid stack is S =
22, 100.

MA232. Determine the number of integers of the form abc + cba, where abc
and cba are three-digit numbers with ac 6= 0.
Originally from Mathematics Competitions Vol. 25, #2 (2012), Heaven and
Earth, heavenly problem 14.
There were 4 submissions, all of them complete and correct. We present the solu-
tion by Catherine Jian.
Since abc = 100a + 10b + c, cba = 100c + 10b + a, we have

abc + cba = 101(a + c) + 20b = 101m + 20b

where m = a + c.
Since b can be 0, 1, . . ., 9, there are 10 ways to choose b. Since a and c are digits
and cannot be 0, m = a + c can be an integer from 2 to 18, i.e. there are 17 ways

Crux Mathematicorum, Vol. 50(2), February 2024

 MathemAttic /63

to choose m. Therefore in total there are 170 possible integers that can be written
as 101m + 20b, or equivalently can be written as abc + cba where abc and cba are
three-digit numbers.

MA233. Determine a polynomial function p(x) with the property that if a

line is drawn and intersects the graph of y = p(x) in two distinct points (a, p(a))
and (b, p(b)), then the y-intercept of the line is ab.
Inspired by John Cook blogpost linked here.
We received 3 submissions, all correct. We present the solution provided by Dim-
itrios Giotas Orfeas.
Let the polynomial be of the form

p(x) = mx2 + nx. (1)

Since the line e intersects the y-axis at (0, ab), it is of the form

y = kx + ab. (2)

Point (a, p(a)) belongs to e, thus

p(a) = ka + ab (3)

Similarly, for point (b, p(b)), we have

p(b) = kb + ab (4)

From equation (1), we get

p(a) = ma2 + na (5)
and
p(b) = mb2 + nb (6)
Combining equations (3) and (5), we get

ka + ab = ma2 + na
k + b = ma + n
k − n = ma − b (7)

Similarly, combining equations (4) and (6), we obtain

kb + ab = mb2 + nb
k + a = mb + n
k − n = mb − a (8)

From equations (7) and (8):

ma − b = mb − a ⇐⇒ m(a − b) = b − a ⇐⇒ m = −1.

Copyright © Canadian Mathematical Society, 2024

64/ MathemAttic

Therefore, equation (7) can be rewritten as

k − n = −a − b ⇒ n = k + a + b.

Thus, the polynomial is in the form

p(x) = −x2 + (k + a + b)x.

As an example, for a = 1, b = 2, k = 3, we have

p(x) = −x2 + (3 + 1 + 2)x = −x2 + 6x.

MA234. Proposed by Ed Barbeau.

Suppose ab and dc are two distinct fractions of positive integers that are both less
than 12 . Prove that the numerator of one of the fractions can be increased by 1 so
that the sum of the two resulting fractions is less than 1.
There were only 2 submissions, both of them complete and correct. We present the
solution by Richard Hess.
Without loss of generality, we can assume ab < dc and let b = 2a + x and d = 2c + y,
where x and y are integers. Then the two fractions are:

a 1 x c 1 y
= − and = − .
2a + x 2 4a + 2x 2c + y 2 4c + 2y

If we add 1 to a, the new sum becomes

2−x y
S1 = 1 + − .
4a + 2x 4c + 2y

If we add 1 to c, the new sum becomes

2−y x
S2 = 1 + − .
4c + 2y 4a + 2x

The largest we can make either sum occurs when x = y = 1 so that

1 1 2(a − c)
S1 = 1 + − =1−
4a + 2 4c + 2 (2a + 1)(2c + 1)

and
1 1 2(c − a)
S2 = 1 + − =1− .
4c + 2 4a + 2 (2a + 1)(2c + 1)
We cannot have a = c for either case where x = y = 1, so that S1 or S2 will always
be < 1 depending upon whether a > c or c > a, respectively.

Crux Mathematicorum, Vol. 50(2), February 2024

 MathemAttic /65

MA235. Proposed by Aravind Mahadevan.

In ∆ABC, cos A cos B + sin A sin B sin C = 1. Find a : b : c, where a, b, and c are
the lengths of sides BC, CA and AB respectively.
There were 6 correct solutions. We present the solution by Miguel Amengual Co-
vas, Bing Jian and Digby Smith, independently.
Since

1 = cos A cos B + sin A sin B sin C ≤ cos A cos B + sin A sin B = cos(A − B) ≤ 1,

the inequality must be an equality. Hence A − B =√0 and sin C = 1. Therefore,

C = 90◦ and A = B = 45◦ , so that a : b : c = 1 : 1 : 2.
Comments from the editor. Richard Hess pointed out that the trigonometric ex-
pression can be written as

cos(A − B) − sin A sin B(1 − sin C).

The minuend is in [−1, 1] and the subtrahend in [0, 1], so the former must be 1
and the latter 0.

Copyright © Canadian Mathematical Society, 2024

66/ Teaching Problems

TEACHING PROBLEMS
No. 25
Ed Barbeau
The Trisection Problem
Often it is the multiple approaches that can be taken or a curious aspect of the
forms of solution that underlie the inclusion of a particular example. In this case,
the contribution addresses one of the more famous unsolved problems in mathemat-
ics. It is shared here also in the hope that some readers will be motivated to submit
ideas to Teaching Problems in future. Interesting problems from your own teach-
ing experiences are welcomed. Send them along to mathemattic@ cms. math. ca .
Some Teaching Problems issues stress pedagogical value of problems and others the
mathematical richness while most blend aspects of both. Here it is the mathematical
side that figures prominently. The experience of working with an unsolved problem
may well be a first for many students, thus enhancing the pedagogical value of the
Trisection Problem.

.................................................................

After students learn about bisecting an angle using straightedge and compasses,
the question of a similar construction for trisecting an angle may come up, and
some students may attempt to find a method. Once I was contacted by a middle
school teacher, one of whose students thought he had succeeded.
The proposed construction was pleasantly simple. Let P OQ be the (acute) angle
to be trisected. From any point A on OP , drop a perpendicular to meet OQ at
B. Construct an equilateral triangle ABC with side AB with O and the vertex C
on opposite sides of AB. Then it is claimed that ∠COB is equal to one third of
∠P OQ. If you check it out with a protractor the method is not bad at all, with
numerical evidence suggesting that the error is within one or two degrees. In fact,
it works for one acute angle; it is not hard to identify and check this angle.
However, there is a pedagogical difficulty here. One could don the mantle of
authority and simply tell the student that it was rigorously proved long ago that
no such method exists. It is more satisfactory to find an explanation that involves
mathematics accessible to the student. I pose two problems for the reader and
suggest solutions for them that I think can be improved upon.
1. Find an argument that the proposed trisection construction is faulty that
involves facts of Euclidean geometry that the student might be expected to know.
The more straightforward the argument the better.
2. Using standard high school mathematics, provide an analysis that identifies
the situations for which the method delivers a trisection.

We are asked to refute a construction that purports to produce a trisection for every

Crux Mathematicorum, Vol. 50(2), February 2024

 Ed Barbeau /67

acute angle. We employ a proof by contradiction: assume that the construction

works for every angle and derive from this a false statement. All we have to do is
to find at least one angle for which it does not work. The following argument will
begin with the assumption that it works for both angles 30◦ and 60◦ and derive
inconsistent conclusions. (In Section 3, you will see how the assumption that it
works for P OQ = 60◦ leads to a contradiction.)
In the diagram below, ∠P OQ = 60◦ and ∠DOB = 30◦ . We will√suppose that
AB = 3, from which we find that BD = 1, AD = 2 and OB = 3. Triangles
ABC and DBE are equilateral, and ∠CBQ = ∠EBQ = 30◦ . Since ACkDE,
CE = AD = 2.
Assuming the method is valid, ∠EOB = 10◦ , so

∠OEB = ∠EBQ − ∠EOB = 20◦ .

Also, by hypothesis, ∠COQ = 20◦ , so ∠COE = 10◦ . Since

∠OCE = ∠OEB − ∠COE = 10◦ ,

triangle COE is isosceles with OE = CE = 2.

Consider triangle OBE, On the one hand, ∠OBE is obtuse. On the other,

OB 2 + BE 2 = 3 + 1 = 4 = OE 2 ,

which can occur only if ∠OBE = 90◦ . Since these two statements are incompati-
ble, the method fails for at least one of 30◦ and 60◦ .

C
D
E
O
B Q

Now look at the general situation of a proper acute angle. In the diagram below,
assume that AB = 2 and OB = t with t > 0. From the diagram, we see that
1
tan ∠COQ = √
t+ 3
and
2
tan ∠P OQ = .
t

Copyright © Canadian Mathematical Society, 2024

68/ Teaching Problems

2 C
1
O √
t B 3 R Q

It can be verified that

√
3t2 + 6t 3 + 8
tan 3∠COQ = √ .
t3 + 3t2 3 + 6t
Therefore
√ √ √
(t3 + 3t2 3 + 6t) · [tan 3∠COQ − tan ∠P OQ] = (3t2 + 6t 3 + 8) − 2(t2 + 3t 3 + 6)
= t2 − 4 = (t − 2)(t + 2).
This vanishes if and only if t = 2 and ∠P OQ = 45◦ . (There is a degenerate
situation when ∠P OQ = 90◦ . Here Q = B and ∠COQ = 30◦ .)
A different assignment of lengths suggested by J. Chris Fisher gives a more trans-
parent √relationship between the tangents
√ of the angles P OQ and COQ.√ Let
AB = 3 and OB = t; then CR = 3/2, BR = 3/2 and tan ∠COQ = 3/(2t+3).
Then
√
3 3(t2 + 3t + 2)
Å 2
3(t + 3t + 2)
ã
tan 3∠COQ = = tan ∠P OQ .
t(2t2 + 9t + 9) 2t2 + 9t + 9

When ∠P OQ = 60◦ , then t = 1 and

√
tan 3∠COQ = (9/10) 3 = (9/10) tan 60◦ .
In this case, the trisection method produces an angle of about 19.1◦ .
Let f (t) = [3(t2 + 3t + 2)]/[2t2 + 9t + 9]. Suppose that ∠P OQ = θ. If t is the
value of OB that corresponds to θ, then 3/t is the value of OB that corresponds
to the complement, 90◦ − θ. Then f (3/t) · f (t) = 1. The function f (t) increases
from 2/3 when t = 0 (and θ = 90◦ ) to 3/2 when t = ∞ (and θ = 0). However, the
effect of the values of f (t) in the accuracy of the trisection when θ is further from
45◦ and f (t) is further from 1 is offset by the fact that when θ is close to 90◦ a
large change in its tangent corresponds to a small change in the angle, and when
θ is small, the error from a true trisection will also be small. For what angle is the
deviation from a proper trisection maximum?

Crux Mathematicorum, Vol. 50(2), February 2024

 OLYMPIAD CORNER /69

OLYMPIAD CORNER
No. 420
The problems featured in this section have appeared in a regional or national mathematical
Olympiad.

Click here to submit solutions, comments and generalizations to any

problem in this section

To facilitate their consideration, solutions should be received by April 15, 2024.

OC666. The squares of a 1×10 board are numbered 1 to 10 in order. Clarissa

and Marissa start from square 1, jump 9 times to the other squares so that they
visit each square once, and end up at square 10. Jumps forward and backward are
allowed. Each jump of Clarissa was for the same distance as the corresponding
jump for Marissa. Does this mean that they both visited squares in the same
order?

OC667. In the convex quadrilateral ABCD, AB and CD are parallel. More-

over, ∠DAC = ∠ABD and ∠CAB = ∠DBC. Is ABCD necessarily a square?

OC668. Consider all 100-digit positive integers such that each digit is 2, 3,
4, 5, 6, 7. How many of these integers are divisible by 2100 ?

OC669. Let M2 (Z) be the set of 2 × 2 matrices with integer entries. Let
A ∈ M2 (Z) such that
A2 + 5I = 0,

where I ∈ M2 (Z) and 0 ∈ M2 (Z) denote the identity and null matrices, respec-
tively. Prove that there exists an invertible matrix C ∈ M2 (Z) with C −1 ∈ M2 (Z)
such that
Å ã Å ã
1 2 0 1
CAC −1 = or CAC −1 = .
−3 −1 −5 0

OC670. Prove that the arithmetic sequence 5, 11, 17, 23, 29, . . . contains
infinitely many primes.

.................................................................

Copyright © Canadian Mathematical Society, 2024

70/ OLYMPIAD CORNER

Les problèmes présentés dans cette section ont déjà été présentés dans le cadre d’une
olympiade mathématique régionale ou nationale.

Cliquez ici afin de soumettre vos solutions, commentaires ou

généralisations aux problèmes proposés dans cette section.

Pour faciliter l’examen des solutions, nous demandons aux lecteurs de les faire parvenir
au plus tard le 15 avril 2024.

OC666. Les cases d’un tableau 1 × 10 sont numérotées de 1 à 10 dans l’ordre.

Clarissa et Marissa partent de la case 1, sautent 9 fois vers les autres cases de
façon à visiter chaque case une fois, et finissent à la case 10. Les sauts vers l’avant
et vers l’arrière sont autorisés. Chaque saut de Clarissa a été effectué sur la même
distance que le saut correspondant de Marissa. Cela signifie-t-il qu’elles ont toutes
deux visité les cases dans le même ordre ?

OC667. Dans le quadrilatère convexe ABCD, AB et CD sont parallèles. De

plus, ∠DAC = ∠ABD et ∠CAB = ∠DBC. Est-ce que ABCD est nécessairement
un carré ?

OC668. Considérons tous les entiers positifs à 100 chiffres dont chacun des
chiffres est 2, 3, 4, 5, 6, ou 7. Combien de ces entiers sont divisibles par 2100 ?

OC669. Soit M2 (Z) l’ensemble des matrices 2 × 2 dont les composantes sont
des entiers. Soit A ∈ M2 (Z) tel que

A2 + 5I = 0,

où I ∈ M2 (Z) et 0 ∈ M2 (Z) désignent respectivement les matrices identité et

nulle. Montrez qu’il existe une matrice inversible C ∈ M2 (Z) avec C −1 ∈ M2 (Z)
telle que Å ã Å ã
−1 1 2 −1 0 1
CAC = ou CAC = .
−3 −1 −5 0

OC670. Montrez que la suite arithmétique 5, 11, 17, 23, 29, . . . contient une
infinité de nombres premiers.

Crux Mathematicorum, Vol. 50(2), February 2024

 OLYMPIAD CORNER /71

OLYMPIAD CORNER
SOLUTIONS
Statements of the problems in this section originally appear in 2023: 49(7), p. 359–360.

OC641. Let a, b, c be integers. Prove that there exists a positive integer n

such that the number n3 + an2 + bn + c is not a perfect square.
Originally Poland Mathematics Olympiad, 8th Problem, First Round 2017.
We received 12 submissions. We present 3 solutions.
Solution 1, by Oliver Geupel.
Let f (x) = x3 + ax2 + bx + c. We will show the stronger result that, for some
integer n in {1, 2, 3, 4}, the number f (n) is congruent to 2 or 3 modulo 4. Since
every perfect square is congruent to either 0 or 1 modulo 4, the required property
follows immediately. Observing that

f (4) ≡ c (mod 4) ,

we need to consider only c ∈ {0, 1}. Next,

f (2) ≡ 2b + c (mod 4)

allows us to restrict attention to b ∈ {0, 2}. The following table surveys all cases,
where all congruences are implicitly modulo 4.

(b, c) (mod 4) a≡0 a≡1 a≡2 a≡3

(0, 0) f (3) ≡ 3 f (1) ≡ 2 f (1) ≡ 3 f (3) ≡ 2
(0, 1) f (1) ≡ 2 f (1) ≡ 3 f (3) ≡ 2 f (3) ≡ 3
(2, 0) f (1) ≡ 3 f (3) ≡ 2 f (3) ≡ 3 f (1) ≡ 2
(2, 1) f (3) ≡ 2 f (3) ≡ 3 f (1) ≡ 2 f (1) ≡ 3

This completes the proof.

Solution 2, by Vivek Mehra.

We prove the stronger result that P (x) = ax3 + bx2 + cx + d cannot be a perfect
square for all positive integers x (where a, b, c, d are given integers with a > 0).
Suppose P (x) = ax3 + bx2 + cx + d is a square for all positive integers x.
Let
P (x) = (F (x))2 G(x), F (x), G(x) ∈ Z[x],
where G(x) is square-free in Z[x]. Then deg G ∈ {1, 3}. Suppose deg G = 1. Then
it is easy to see that G(x) is not a square for infinitely many positive integers
x. Also (F (x))2 = 0 for only finitely many x. So G(y) is not a square for some

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72/ OLYMPIAD CORNER

positive integer y such that F (y) 6= 0 and then P (y) = (F (y))2 G(y) is not a perfect
square leading to a contradiction. So deg G = 3. But by the argument given in
solution to Crux problem OC124 (Vol 40 no. 5, p. 198-199), we have that deg G
cannot be ≥ 2 and so deg G 6= 3. So we get that P (x) cannot be a square for all
positive integers x.

Solution 3, by UCLan Cyprus Problem Solving Group.

This is known in an even more general setting: If P (x) ∈ Z[x] satisfies that P (n)
is a perfect k-th power whenever n ∈ N, then P (x) = Q(x)k for some Q(x) ∈ Z[x].
(See e.g. Part Eight, Problem 114 of G. Pólya and G. Szegő, Problems and theorems
in analysis Vol. II, 1976.)

We provide a proof in our particular instance. It doesn’t generalise but we believe

it is simpler than the more general proof.

Let P (x) = x3 + ax2 + bx + c. We claim that there is an integer n and a prime

p ≥ 5 such that p | Q(n).

If c = 0, then this is possible by taking any p ≥ 5 and n = p. If c 6= 0, then

P (6cm) = c[6Q(m) + 1]

for some cubic polynomial Q(x). So the claim holds by taking n = 6cm for some
m such that Q(m) 6= 0 and any prime p with p | 6Q(m) + 1.

Pick a prime p as in the above paragraph. If p2 - P (n), then we are done so assume
that p2 | P (n). Then

P (n + p) − P (n) ≡ p(3n2 + 2an + b) ≡ pP 0 (n) mod p2 .

If p - P 0 (n), then p | P (n + p) but p2 - P (n + p). This implies that P (n + p) is not

a perfect square and we are done.

So we may assume that p | P (n) and p | P 0 (n). This implies that over Fp we can
write
P (x) = (x − n)2 (x − m)

for some integer m. Let k 6≡ (n − m) (mod p) be a number which is not a perfect

square modulo p. Such a number exists because every p ≥ 5 has at least two
non-quadratic residues. Then

P (k + m) ≡ k(k + m − n)2 (mod p)

so it is not a perfect square modulo p. But then P (k + m) is not a perfect square.

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 OLYMPIAD CORNER /73

OC642. Determine if there exist positive integers n and k such that

n
11k − n

is the square of an integer.

Originally Czech-Slovakia Math Olympiad, 4th Problem, Category A, Regional
Round 2018.
We received 15 correct submissions. We present 2 solutions.
Solution 1, by UCLan Cyprus Problem Solving Group.
n
There is no such number. Indeed, assume by contradiction that = m2 for
11k−n
11k m2
some positive integer m. Then n = .
m2 + 1
Pick any prime p such that p | (m2 + 1). Then p - m2 , so p | 11k and therefore
p = 11. Then m2 ≡ −1 (mod 11) which is a contradiction as −1 is not a perfect
square modulo 11 (as 11 ≡ 3 (mod 4)).

Solution 2, by Oliver Geupel.

We show that there are no such numbers. Suppose that k, m, and n
are positive
integers with the property that n = 11k − n m2 . Then m2 + 1 n = 11k m2 .


Since m2 and m2 + 1 are relatively prime, we deduce that m2 + 1 is a divisor of
11k . Hence, there is a positive integer q such that m2 + 1 = 11q . If q were even,
say, q = 2s, we would obtain (11s − m) (11s + m) = 1, which is impossible. Thus
q is odd. With the notation q = 2s + 1, we find that
2s
X
m2 = 11q − 1 = 10 11j ,
j=0

so that m is divisible by 10. Let m = 10t. We conclude

2s
X
0 ≡ 10t2 ≡ 11j ≡ 2s + 1 (mod 10) ,
j=0

a contradiction.
Editor’s Comment. Roy Barbara and Konstantine Zelator proved the more general
case: if n, k ∈ N and p is a prime, p ≡ 3 (mod 4), then pkn−n is not the square
of an integer. Giuseppe Fera also made the same remark. The proof is done by
contradiction and by using some considerations about the prime factorization of an
integer. We leave the details to the reader. Also, Roy Barbara proved a stronger
result: under the same hypothesis, the number pkn−n is not the square of a rational
number.

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74/ OLYMPIAD CORNER

OC643. Find the smallest natural number n such that for every 3-colouring
of the numbers 1, 2, . . . , n there are two (different) numbers of the same colour
such that their positive difference is a perfect square.
Originally Czech-Slovakia Math Olympiad, 6th Problem, Category A, Final Round
2018.
We received 4 solutions, of which 3 were correct and complete. We present the
solution by UCLan Cyprus Problem Solving Group.
We will show that the smallest such n is n = 29.
Note first that n ≥ 29 since in the following 3-colouring of the first 28 numbers, no
two numbers of the same colour differ by a perfect square: colour the numbers in
{1, 4, 6, 9, 12, 14, 19, 24, 27} red, the ones in {2, 5, 7, 10, 15, 17, 20, 22, 25, 28} green,
and the ones in {3, 8, 11, 13, 16, 18, 21, 23, 26} blue.
To see that we cannot colour the first 29 numbers in such a way, observe that 10
and 17 must be similarly coloured. Indeed if they are differently coloured, then 1
must take a different colour to both, and so must 26. But then 1 and 26 would
have the same colour, a contradiction. Similarly, the following pairs of numbers
should be similarly coloured (11, 18), (12, 19), (13, 20). (For the pair (13, 20) we
look at the numbers 4 and 29.)
Suppose without loss of generality that 10, 17 are coloured red. Then 11, 18 must
have a different colour, say green. Then 12, 19 cannot be red (since 19 − 10 = 9),
but they also cannot be green (since 12 − 11 = 1). Say that they are both coloured
blue. But then there is no available colour for 13 and 20. They cannot be red
as 17 − 13 = 4, they cannot be green as 20 − 11 = 9 and they cannot be blue as
20 − 19 = 1.

OC644. In a 2018 × 2018 chessboard, some of the cells are painted white, the
rest are black. It is known that from this chessboard one can cut out a 10 × 10
square, all cells of which are white, and a 10 × 10 square, all cells of which are
black. What is the smallest d for which it can be guaranteed that a 10 × 10 square
can be cut out of it, in which the number of black and white cells differs by no
more than d?
Originally Moscow Math Olympiad, 2nd Problem, Grade 10, 2018.
We received 2 correct submissions. We present both approaches.
Solution 1, by UCLan Cyprus Problem Solving Group.
For each 10 × 10 square S, let WS be the number of white squares in S and BS
the number of black squares in S. Let dS = WS − BS . We know that there is a
square S1 with dS1 = 100 and a square S2 with dS2 = −100.
Create a graph with vertices the 10 × 10 squares, where two such squares are
neighbours if and only if they intersect in a 9 × 10 or 10 × 9 rectangle. In this
graph there is a path from S1 to S2 .

Crux Mathematicorum, Vol. 50(2), February 2024

 OLYMPIAD CORNER /75

So we can find a sequence of 10 × 10 squares, T1 , T2 , . . . , Tk such that the sequence

dT1 , dT2 , . . . , dTk starts with 100, ends with −100 and consecutive elements differ
by at most 20. So one of these elements must belong in the set {−10, −9, . . . , 9, 10}.
So we can find a 10 × 10 square in which the number of white and black squares
differ by at most 10.
We can not improve on this. Consider for example the following ‘diagonal colour-
ing’ of the square: Colour the cell (i, j) (where 1 ≤ i, j ≤ 100) white if i + j ≤ 100
and black otherwise. Then every 10 × 10 square is also diagonally coloured. The
diagonals have sizes 1, 2, . . . , 9, 10, 9, . . . , 2, 1 with all cells in a diagonal having the
same colour. The first few (maybe none) of these diagonals are all coloured white
and the rest black. The colour which appears in the diagonal of size 10 occurs at
least 10 more times than the other colour in this square.

Solution 2, by Oliver Geupel.

We prove that the smallest d is equal to 10.
Assume that we have a frame that encloses a 10×10 square. A move is a translation
of the frame on the board by one column to the left or right or by one row up or
down. Starting with the frame on a 10 × 10 white square, an appropriate finite
sequence of moves, say, of m moves, will bring the frame to enclosing a 10×10 black
square. Consider the corresponding sequence b0 , b1 , b2 , . . . , bm where b0 = 0 and,
for 1 ≤ k ≤ m, bk is the number of black cells inside the frame after the k th move.
In each move, any 10 cells are replaced by other 10 cells, so that |bk − bk−1 | ≤ 10
for 1 ≤ k ≤ m. Since b0 = 0 and bm = 100, there exists an index k with the
property that 45 ≤ bk ≤ 55. It follows that in the framed 10 × 10 square after the
k th move the numbers of black and white cells differ by at most 10.
We finish the proof by paiting the cells of a board so that the smallest number d
with the desired property is equal to 10. Let us paint the 10 left columns all white
and call them the white region. Further, paint the 2007 right columns all black
and call them the black region. Finally, paint the cells of the column between the
black and the white region alternating black and white. Then, for every 10 × 10
square which is fully contained in the white or black region, the numbers of black
and white cells differ by 100. A square that is not completely contained in one
of the regions, consists of some n white columns, some 9 − n black columns and
an alternatingly coloured border column, resulting in 10n + 5 white and 95 − 10n
black cells. For n ∈ {4, 5} we reach the minimum d = 10.

OC645. Prove that if a, b, c ≥ 0 and a + b + c = 3, then

a b c 1 1 1
+ + ≥ + + .
1+b 1+c 1+a 1+a 1+b 1+c

Originally Romania Math Olympiad, 2nd Problem, Grade 9, Final Round 2018.
We received 15 submissions. We present the solution by Theo Koupelis.

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76/ OLYMPIAD CORNER

The desired inequality is equivalent to

a−1 b−1 c−1
+ + ≥ 0,
1+b 1+c 1+a
or after clearing denominators,

(a2 − 1)(c + 1) + (b2 − 1)(a + 1) + (c2 − 1)(b + 1) ≥ 0 ⇐⇒

a2 c + b2 a + c2 b + [a2 + b2 + c2 − (a + b + c) − 3] ≥ 0.

But

a2 + b2 + c2 = (a + b + c)2 − 2(ab + bc + ca) = 3 + 2(a + b + c) − 2(ab + bc + ca).

Substituting, we get the equivalent inequality

(a2 c + c) + (b2 a + a) + (c2 b + b) ≥ 2ac + 2ab + 2bc,

which is obvious by AM-GM. Equality occurs when a = b = c = 1.

Crux Mathematicorum, Vol. 50(2), February 2024

 Yagub Aliyev /77

Reading a Math Book

Solutions to No. 2
Yagub Aliyev
The statements of the problems in this section originally appear in 2023: 49(5),
p. 266-273. The problems were selected from [1].

S6. Prove that if the three vertices A, A0 , A00 of a changing triangle are moving
along three fixed and concurrent lines u, u0 , u00 , respectively, and its two sides
A0 A00 and A00 A are rotating about two fixed points O and O0 , respectively, then
the third side AA0 is also rotating about a fixed point O00 , which is on the line
OO0 . [1, Problem 51]

Solution.

The triangles obtained by moving the vertices of 4AA0 A00 along the lines u, u0 , u00
are in central perspective from the point of intersection of u, u0 , u00 . Desargues’
theorem then guarantees that such triangles are in axial perspective from the line
OO0 .

S7. Prove that the lines joining a point on a circle with the endpoints of a chord
of the circle divide the diameter perpendicular to the chord harmonically. ([1],
Problem 32).

Solution 1.

Let AB be the given chord, CD the diameter and E the given point. Let lines AE
and BE intersect line CD at points G and F . The result follows from the easily
provable fact that EC and ED are internal and external bisectors of ∠GEF :

EG CG EG DG GC FC
= and = so = .
EF CF EF DF GD FD

Solution 2, by Oliver Geupel.

Denote the given circle, the point on it, the chord, and the perpendicular diameter
by Γ , P , AB, and CD, respectively. Suppose that the lines AP and BP meet the
line CD at points Q and R, respectively. Without loss of generality assume that
the points A and P lie on a common semicircle CD of Γ . Then Q is an exterior
point and R an interior point of the diameter CD.

Consider the problem in the complex plane where Γ is the unit circle and where
a lowercase letter denotes the affix of a point designated by the corresponding
uppercase letter (so, the affix of the point P = (a, b) is the complex number
p = a + bi). We may assume that b = a = 1/a, c = 1, and d = −1. It is known

Copyright © Canadian Mathematical Society, 2024

78/ Reading a Math Book: Solutions to No. 2

that the intersection of two arbitrary chords XY and ZW has affix

xy(z + w) − zw(x + y)
. (1)
xy − zw
Therefore, since AP ∩ CD = Q and BP ∩ CD = R,
ap(c + d) − cd(a + p) a+p
q= =
ap − cd ap + 1
and
bp(c + d) − cd(b + p) b+p 1 + ap 1
r= = = = .
bp − cd bp + 1 p+a q
Hence,
1
CQ q−1 1− q 1−r CR
= = 1 = = ,
DQ q+1 1+ q
r+1 DR
which completes the proof.

S8. Given are a fixed circle and two fixed points A and B on it. On the same circle
two arbitrary points C and D are chosen. Let M = AC ∩ BD and N = AD ∩ BC.
Prove that as the points C and D change, the line M N passes through a fixed
point of the plane. [1, Problem 162]
Solution by Oliver Geupel.
We prove that M N passes through the intersection T of the tangents at A and
B. Consider the problem in the complex plane where A, B, C, and D lie on the
unit circle. Let a, b, c, d, m, n, and t be the affixes of A, B, C, D, M , N , and T ,
respectively. Using equation (1) from the previous problem, we get
2ab f (c, d) f (d, c)
t= , m= , n= ,
a+b g(c, d) g(d, c)
where
f (x, y) = ax(b + y) − by(a + x)
and
g(x, y) = ax − by.
The points M , N , and T are collinear if

(t − m)(t − n) = (t − n)(t − m). (2)

For a, b, c and d lie on the unit circle, we have

1 1 1 1
a= , b= , c= , d= ,
a b c d
which gives
2 h(c, d) h(d, c)
t= , m= , n= ,
a+b g(c, d) g(d, c)

Crux Mathematicorum, Vol. 50(2), February 2024

 Yagub Aliyev /79

where
h(x, y) = a − b + x − y.
The collinearity condition thus simplifies to

F (c, d)G(c, d) = F (d, c)G(d, c), (3)

where
F (x, y) = 2ab · g(x, y) − (a + b) · f (x, y)
and
G(x, y) = 2g(y, x) − (a + b) · h(y, x).
With a little algebra, we easily obtain

F (x, y) = (a − b) (ab(x + y) − (a + b)xy)

and
G(x, y) = (a − b)(x + y − a − b).
This shows that the polynomials F and G are symmetric in the variables x and y.
Hence (1) holds, which completes the proof.
Note that the problem can also be solved using Pascal’s theorem by applying it to
the hexagon AACBBD.

S9. Let the tangent lines of the circumcircle of 4ABC at the points B and C
intersect at point D. Through D draw the line that is parallel to the tangent
line of the circle at A. This line intersects lines AB and AC at points E and F ,
respectively. Show that D is the midpoint of the segment EF . [1, Problem 159]
Solution.

Denote the intersection points of the tangent of the circumcircle at point A with
the tangents at B and C by G and H, respectively. It is obvious that GA = GB,
HA = HC, and DB = DC. By similarity of triangles BDE and BGA, and of
triangles CDF and CHA, we obtain
BD DE
=
BG GA

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80/ Reading a Math Book: Solutions to No. 2

and
CD DF
= .
CH HA
This means that BD = DE and CD = DF , whence DE = DF .

S10. Given are a triangle ABC and its incircle touching the side BC at the
point D. From a point A1 on the side BC, a second tangent A1 A2 of the incircle
is drawn and the tangency point A2 is connected with A by a line. The line AA2
intersects the side BC at D1 . Prove that
Å ã2
DB D1 B A1 B
· = .
DC D1 C A1 C

[1, Problem 35]

Solution.

Denote AB = c = x + y, AC = b = x + z, and BC = a = y + z as in the picture.

Denote also CA1 = u, CD1 = t, and CB1 = v. Then DA1 = z − u, A1 D1 = u − t,
and EB1 = z − v. By Menelaus’ theorem,

t z−u z−v+x
· · = 1.
u−t z−v z+x
By solving this for t, we find:

u vx + zv − zx − z 2


t=− . (4)
uv − ux − uz − vx − 2zv + 2zx + 2z 2
On the other hand, it is known that

(a + b + c)uv − 2ab(u + v) + ab(a + b − c) = 0,

Crux Mathematicorum, Vol. 50(2), February 2024

 Yagub Aliyev /81

(see [3, p. 101, Exercise 47]) which can be written as

(x + y + z)uv − (y + z)(x + z)(u + v) + (y + z)(x + z)z = 0.

By solving this for x, we find

uvy + uvz − uyz − u z 2 − vyz − v z 2 + z 2 y + z 3

x=− . (5)
uv − yu − uz − yv − zv + zy + z 2

By substituting (5) in (4), and using computer assistance we can find

y u2
t= ,
(u − z)2 + yz

which we can substitute into the claimed identity

2
y (y + z − t) (y + z − u)
· = ,
z t u2
to check that it is correct.

References
[1] M.P. Chernyaev, Problems in synthetic geometry, Rostov University, Rostov-
on-Don, 1961 (in Russian).
[2] Y. Aliyev, Reading a Math Book: No. 2 (M.P. Chernyaev: Problems in syn-
thetic geometry), Crux Mathematicorum, 49(5), May 2023, 266–272.
[3] M.B. Balk, V.G. Boltyanskiy, Geometriya mass, Library Kvant 61, 1987 (in
Russian).

Copyright © Canadian Mathematical Society, 2024

82/ Problems

PROBLEMS
Click here to submit problems proposals as well as solutions, comments
and generalizations to any problem in this section.

To facilitate their consideration, solutions should be received by April 15, 2024.

4911. Proposed by Mihaela Berindeanu, modified by the Editorial Board.

Given a triangle ABC with ∠BAC = 60◦ , let P denote one of the points where
its circumcircle intersects the perpendicular bisector of AC, and T denote the foot
of the perpendicular from P to the bisector of ∠BAC. Prove that P T is tangent
to the nine-point circle of ∆ABC at T .

4912. Proposed by Michel Bataille.

Let P be a point inside an equilateral triangle ABC with side a. Prove that
P A, P B and P C are the sides of a triangle T and that T has an angle of 60◦ if
and only one of its medians has length a2 .

4913. Proposed by Albert Natian.

Suppose the continuous function f satisfies the integral equation

xf (7)
tx2
Z Å ã
f dt = 3f (7) x4 .
0 f (7)

Find f (7).

4914. Proposed by Ivan Hadinata.

Let R≥0 be the set of all non-negative real numbers. Find all possible monotoni-
cally increasing f : R≥0 → R≥0 satisfying

f (x2 + y + 1) = xf (x) + f (y) + 1, ∀x, y ∈ R≥0 .

4915. Proposed by Michel Bataille.

∞
X (−1)k+1
Let Sn = , where n is a nonnegative integer. Find real numbers
k(k + n + 1)
k=1
a, b, c such that lim n3 Sn − (an2 + bn + c) = 0.


n→∞

Crux Mathematicorum, Vol. 50(2), February 2024

 Problems /83

4916. Proposed by Arsalan Wares.

Equilateral triangle ABC is split into 5 isosceles trapezoids and a smaller equilat-
eral triangle, all of the same area. One of the side lengths of the shaded trapezoid
is 10. Determine the exact length of AB.

4917. Proposed by Pericles Papadopoulos.

Let D, E and F be the points of contact of the incircle of a triangle ABC with
the sides BC, AC and AB, respectively. Let S, T and U be the orthocenters of the
triangles EAF, F BD and DCE, respectively. Prove that SD, T E and U F concur
at a point.

4918. Proposed by Yagub Aliyev.

2
λx
Let L = limλ→+∞ R b .
a
λt2 dt
a) Show that if 0 ≤ a ≤ x < b, then L = 0.
b) Show that if 0 ≤ a < x = b, then L = +∞.

4919. Proposed by Daniel Sitaru.

If A, B ∈ M6 (R) are matrices such that
A2 + B 2 = AB + A + B − I6 ,
then
det(BA − AB) ≥ 0

Copyright © Canadian Mathematical Society, 2024

84/ Problems

4920. Proposed by Ángel Plaza.

Z 1
log(1 + xk + x2k + · · · + xnk )
If k > 1 and n ∈ N, evaluate dx.
0 x

.................................................................

Cliquez ici afin de proposer de nouveaux problèmes, de même que pour

offrir des solutions, commentaires ou généralisations aux problèmes
proposés dans cette section.

Pour faciliter l’examen des solutions, nous demandons aux lecteurs de les faire parvenir
au plus tard le 15 avril 2024.

4911. Proposée par Mihaela Berindeanu, modifié par le comité de rédaction.

Soit ABC un triangle tel que ∠BAC = 60◦ et soient P un des points d’intersection
du cercle circonscrit et de la bissectrice orthogonale de AC, puis T le point d’arrivée
de la perpendiculaire de P vers la bissectrice de ∠BAC. Démontrer que P T est
tangent en T au cercle des neuf points de ∆ABC.

4912. Proposée par Michel Bataille.

Soit P un point à l’intérieur d’un triangle ABC équilatéral de côté a. Démontrer
que P A, P B and P C sont les côtés d’un triangle T , puis que T a un angle de 60◦
si et seulement si une de ses médianes est de longueur a2 .

4913. Proposée par Albert Natian.

Supposer que la fonction continue f est solution de l’équation intégrale
xf (7)
tx2
Z Å ã
f dt = 3f (7) x4 .
0 f (7)

Déterminer f (7).

4914. Proposée par Ivan Hadinata.

Soit R≥0 l’ensemble des nombres réels non négatifs. Déterminer toutes les fonc-
tions non décroissantes f : R≥0 → R≥0 telles que

f (x2 + y + 1) = xf (x) + f (y) + 1, ∀x, y ∈ R≥0 .

Crux Mathematicorum, Vol. 50(2), February 2024

 Problems /85

4915. Proposée par Michel Bataille.

∞
X (−1)k+1
Soit Sn = , où n est un entier non négatif. Déterminer des nom-
k(k + n + 1)
k=1
bres réels a, b, c tels que lim n3 Sn − (an2 + bn + c) = 0.


n→∞

4916. Proposée par Arsalan Wares.

Le triangle équilatéral ABC est divisé en 5 trapézoı̈des isocèles et un triangle
équilatéral, tous de même surface. Un côté d’un des trapézoı̈des est de longueur
10, tel qu’indiqué. Déterminer la valeur exacte de la longueur de AB.

4917. Proposée par Pericles Papadopoulos.

Soient D, E et F les points de contact du cercle inscrit du triangle ABC avec ses
côtés BC, AC et AB, respectivement, puis soient S, T et U les orthocentres des
triangles EAF , F BD et DCE, respectivement. Démontrer que SD, T E et U F
sont concourantes.

4918. Proposée par Yagub Aliyev.

2
λx
Soit L = limλ→+∞ R b 2 .
a
λt dt
a) Démontrer que si 0 ≤ a ≤ x < b, alors L = 0.
b) Démontrer que si 0 ≤ a < x = b, alors L = +∞.

Copyright © Canadian Mathematical Society, 2024

86/ Problems

4919. Proposée par Daniel Sitaru.

Si A, B ∈ M6 (R) sont des matrices telles que

A2 + B 2 = AB + A + B − I6 ,

alors
det(BA − AB) ≥ 0

4920. Proposée par Ángel Plaza.

Z 1
log(1 + xk + x2k + · · · + xnk )
Si k > 1 et n ∈ N, évaluer dx.
0 x

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 Solutions /87

SOLUTIONS
No problem is ever permanently closed. The editor is always pleased to consider for
publication new solutions or new insights on past problems.
Statements of the problems in this section originally appear in 2023: 49(7), p. 375–380.

4861. Proposed by Pericles Papadopoulos.

Let I be the incenter of a triangle ABC and let D, E and F be the points of
contact of the incircle of the triangle with the side BC, AC and AB respectively.
The circle AIB meets the sides BC and AC at points K and P respectively; the
circle AIC meets the sides BC and AB at points L and T respectively; the circle
BIC meets the sides AC and AB at points Q and S, respectively. Prove the
following:
• KL + P Q + ST = AB + BC + AC
• Points D, E and F are the midpoints of KL, P Q and ST respectively.

All 16 submissions that we received are correct; we provide a composite of the

similar solutions from Michal Adamaszek, Oliver Geupel, Antoine Mhanna and
the UCLan Cyprus Problem Solving Group.
As usual, let r be the radius of the incircle of ∆ABC, a = BC, b = CA, c = AB,
and 2s = a+b+c. Let’s begin with an outline of the argument. Since A, B, K, I are
concyclic and D is on the extension of the ray BK, then ∠DKI = ∠BAI = A2 .
Since also ID = IF = r, then the right-angled triangles IKD and IAF are

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88/ Solutions

congruent. We therefore have KD = AF , where AF = s − a is the length of the

tangent from the vertex A to the incircle. We similarly have

KD = LD = AF = AE = s − a
QE = P E = BD = BF = s − b
T F = SF = CE = CD = s − c.

Note that KL = KD + DL, P Q = P E + EQ, and ST = SF + F T . Both parts of

the problem now follow immediately.
For a detailed proof, one must observe that since the incenter I is always inside
the triangle, any circle through two vertices and I must intersect the sides through
the third vertex at points interior to one side and exterior to the other (except,
of course, when the circle is tangent to those two sides). Any informal argument
would therefore depend on how the five relevant points are arranged on each cir-
cle. Geupel avoided the difficulty through the use of directed angles. Here is his
argument:
We use the notation ∠(p, q) to denote the directed angle from the line p to the
line q, taken modulo 180◦ . Because A, B, I, and K are concyclic and also A, C,
I, and L are concyclic, we can calculate

∠(LK, IK) = ∠(BK, IK) = ∠(BA, IA) = ∠(IA, CA) = ∠(IL, CL) = ∠(IL, KL).

Hence, triangle IKL is isosceles with ∠IKL = ∠ILK = A/2, and the foot D of its
altitude DI is the midpoint of KL. Analogously, E and F are the midpoints of P Q
and ST , respectively. Triangles AEI and KDI are congruent by comparison of two
angles and EI = DI. Therefore, KD = AE = s − a and KL = 2KD = 2(s − a).
With the similar identities P Q = 2(s − b) and ST = 2(s − c), we finally conclude
that

KL + P Q + ST = 2(s − a + s − b + s − c) = 2s = AB + BC + AC.

4862. Proposed by Michel Bataille.

Let m be a nonnegative integer. Find
n Ç åÇ å
1 X m+k m+n+1
lim .
n→∞ 2n nm k n−k
k=0

We received 10 submissions, 8 of which were correct. We present the joint solution

of Ulrich Abel and Vitaliy Kushnirevych.
By taking the Cauchy product, we see that the sum
n Ç åÇ å
X m+k m+n+1
Sm (n) :=
k n−k
k=0

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 Solutions /89

is the coefficient of xn in the power series expansion of

∞ Ç å ∞ Ç å
X m+k k X m+n+1 k −m−1 m+n+1
x · x = (1 − x) · (1 + x)
k k
k=0 k=0

where both series are absolutely convergent for |x| < 1. Using 1 + x = 2 + (x − 1)
and the binomial formula, this coefficient is
m+n+1 Ç å
1 dn m+1
X m+n+1 k n−k
Sm (n) = (−1) 2 (x − 1)
n! dxn x=0 k
k=0
m+n+1 Ç å
1 m+1
X m+n+1 k n −k
= (−1) 2 (n − k) (−1)
n! k
k=0

with the falling factorial


n!
 (k = 0),
n
(n − k) := (n − k) (n − 1 − k) . . . (1 − k) = 0 (1 ≤ k ≤ n),
(−1)n
 (k−1)!
(k > n).
(k−1−n)!

After some manipulations (namely, pulling out the k = 0 term and writing k =
` + n + 1 in the tail) we obtain
m Ç åÇ å
m+1 m−` m + n + 1 `+n `
X
Sm (n) = (−1) + 2n+1 (−1) 2.
m−` `
`=0

Observing that m+n+1

`+n

m−` ` is a polynomial of degree m in the variable n with
leading coefficient 1/ ((m − `)!`!) we infer that
m Ç å
1 2 X m−` m 2 m 2
lim n m Sm (n) = (−1) 2` = (2 − 1) = .
n→∞ 2 n m! ` m! m!
`=0

Editor’s Comments. We chose to feature this approach because it tells us that

2n+1
Sm (n) = (−1)m+1 + Pm (n)
m!
where
m Ç å
m−` m (n + m + 1)! `
X
Pm (n) = (−1) 2
` (n + ` + 1)n!
`=0
is a monic polynomial of degree m with integer coefficients. For example,
P0 (n) = 1
P1 (n) = n
P2 (n) = n2 + n + 2
P3 (n) = n3 + 3n2 + 8n
P4 (n) = n4 + 6n3 + 23n2 + 18n + 24
P5 (n) = n5 + 10n4 + 55n3 + 110n2 + 184n

Copyright © Canadian Mathematical Society, 2024

90/ Solutions

and so on. The majority of solvers rewrote Sm (n) in terms of the quantity
n Ç å Z 1
X n 1
In,m = = xm (1 + x)n dx
k m+k+1 0
k=0

and continued their analyses from there. A few others found the generating-
function identity
∞ ∞ Xn Ç å
X
n 1 X m + i i n
Sm (n)x = = 2 x
n=0
(1 − x)(1 − 2x)m+1 n=0 i=0
i
which lends itself very well to elementary reasoning.

4863. Proposed by Mihaela Berindeanu, modified by the Editorial Board.

In a parallelogram ABCD, let E be the point where the diagonal BD is tangent
to the incircle of ∆ABD. If r1 and r2 are the inradii of the triangles DEC and
BEC, prove that rr21 = DE
EB .

We received solutions from 17 solvers, all correct. The following is by Miguel

Amengual Covas.
Solution. Let s be the semiperimeter of 4ABD and let the two pairs of equal
sides BC, DA and AB, CD of ABCD be labeled x and y, respectively.
In 4ABD, then
BE = s − x ED = s − y,
implying that triangles BEC and DEC have equal perimeters (being the common
perimeter = s + CE). Since the area of a triangle is equal to the product of the
inradius and the semiperimeter, the areas of triangles with equal perimeters are
proportional to the inradii of the triangles, hence
[DEC] r1
= .
[BEC] r2

A x D

s−y

s−x

B x C

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 Solutions /91

Now, the areas of triangles with equal altitudes are proportional to the bases of
the triangles, hence
[DEC] DE
= ,
[BEC] EB
and therefore
DE r1
= ,
EB r2
as desired.

4864. Proposed by Goran Conar.

Let a, b, c be side-lengths of an arbitrary three-dimensional box, and D the length
of its main diagonal. Prove

√ √ √ (a + b + c)2 D2
1+a+ 1+b+ 1+c≥ · 1+ .
D2 a+b+c

When does the equality occur?

We received 7 submissions, all correct. We presents the solution by Theo Koupelis.
Using Minkowski’s inequality we get
√ √ √ » √ √ √
1+a+ 1+b+ 1+c≥ 9 + ( a + b + c)2 .

With D2 = a2 + b2 + c2 , it is sufficient then to show that

î √ √ √ ó
(a2 + b2 + c2 )2 9 + ( a + b + c)2 ≥ (a + b + c)4 + (a2 + b2 + c2 )(a + b + c)3 .

This is obvious because

3(a2 + b2 + c2 ) ≥ (a + b + c)2

by AM-GM, and
√ √ √
(a2 + b2 + c2 )( a + b + c)2 ≥ (a + b + c)3

by Hölder’s inequality. Equality occurs when a = b = c.

4865. Proposed by George Apostolopoulos.

Let ABC be an acute triangle with inradius r and circumradius R. Prove that

(sec A)cos A + (sec B)cos B + (sec C)cos C 5R − r

< .
sec A + sec B + sec C 12r

We receive 10 submissions, of which 2 were incomplete or incorrect. We present

2 solutions.

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92/ Solutions

Solution 1, by Theo Koupelis, slightly modified by the Editorial Board.

Let f (x) = 1/xx , where x > 0.
We have f 0 (x) = −(1 + ln x)/xx and f 00 (x) = −[1 − x(ln x + 1)2 ]/xx+1 . Thus, the
function f (x) is concave in (0, 1), with a maximum equal to e1/e at x = e−1 , and
continuously decreasing, and convex for x > 1, with f (1) = 1.
Using Jensen’s inequality we get

(sec A)cos A + (sec B)cos B + (sec C)cos C ≤

ã(cos A+cos B+cos C)/3
3
Å
3· .
cos A + cos B + cos C

As e1/e ≈ 1.444667 < 32 ,

9
(sec A)cos A + (sec B)cos B + (sec C)cos C ≤ 3 · e1/e < .
2
But
A+B A−B A+B
Å ã
cos A + cos B + cos C = 1 + 2 cos · cos − cos ,
2 2 2

and thus 1 < cos A + cos B + cos C ≤ 23 . The maximum occurs when the triangle
is equilateral and the minimum is approached when two of the angles tend to π2 − .
Also, from Cauchy-Schwarz inequality we get
9
sec A + sec B + sec C ≥ ≥ 6.
cos A + cos B + cos C
Therefore,

(sec A)cos A + (sec B)cos B + (sec C)cos C 9 5R − r

< ≤ ,
sec A + sec B + sec C 12 12r
where in the last step we used Euler’s inequality R ≥ 2r.

Solution 2, by Marie-Nicole Gras.

We put
1 1 1
F = + + ·
cos Acos A cos B cos B cos C cos C
We consider the function f (x) = xx = exp x log(x) , where x = cos A; since


4ABC is acute, we have 0 < x < 1.
We have f 0 (x) = log(x) + 1 f (x); for 0 < x < e−1 , then f 0 (x) < 0, f 0 (e−1 ) = 0


−1
and for x > e−1 , then f 0 (x) > 0; we deduce that xx ≥ (e−1 )e , and
1 −1
≤ ee ≈ 1.444667.
xx

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 Solutions /93

We deduce that F < 4.5.

The Cauchy-Schwarz inequality implies

 1 1 1 
cos A + cos B + cos C + + ≥ 9;
cos A cos B cos C
it follows
F cos A + cos B + cos C cos A + cos B + cos C
1 1 1 < 4.5 × = ·
cos A + cos B + cos C
9 2
R+r
However, it is well known that cos A + cos B + cos C = R ; then, to prove the
required inequality, it is enough to prove that
R+r 5R − r
≤ · (1)
2R 12r
Let H = 5R2 − 7Rr − 6r2 ; then H = (5R + 3r)(R − 2r), and Euler’s inequality
implies F ≥ 0, whence (1) and the result.

4866. Proposed by Ivan Hadinata.

Find all functions f : R → R such that the equation

f (xy + f (f (y))) = xf (y) + y

holds for all real numbers x and y.

We received 21 submissions and they were all complete and correct. We present
the following two solutions by the majority of solvers.
We show that the only solution to the given functional equation is f (x) = x. By
taking x = 0, we have f (f (f (y))) = y for each y ∈ R, and thus f is bijective. By
taking y = 0, we have 0 = f (f (f (0)) = xf (0) for all x ∈ R, and thus f (0) = 0. To
show that f (x) = x, there are multiple different ways.
Solution 1.
For each y 6= 0, by taking x = −f (f (y))/y, we get

f (f (y))f (y)
0 = f (0) = − +y
y
which leads to
f (f (y))f (y) = y 2 . (1)
Since f is bijective, by replacing y with f (y) in equation (1), we get

yf (f (y)) = f (f (f (y)))f (f (y)) = f (y)2 (2)

for each y 6= 0. Taking the ratio between equation (1) and equation (2), for each
y 6= 0, we have f (y)/y = y 2 /f (y)2 , which implies that f (y) = y.

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94/ Solutions

Solution 2.
Choosing y = 1 leads to f (x + c) = xf (1) + 1, where c = f (f (1)). Replacing x
with x − c, we obtain
f (x) = (x − c) f (1) + 1.
It follows that 0 = f (0) = −cf (1) + 1. Hence, f (x) = ax, where a = f (1).
Inserting this into the functional equation, we get

a xy + a2 y = axy + y.

It is easy to derive that a = 1. We conclude that f (x) = x.

4867. Proposed by Thanos Kalogerakis.

Consider a triangle ABC with |AC| > |BC| > |AB| and let M be the midpoint
of BC. Let K, L and N be points on the sides of ABC (see the figure) such that
the points K, L, M, N divide the perimeter of ABC into 4 equal parts. Prove that
KM bisects LN .

We received 17 submissions, all of which are correct and complete. We present here
two solutions with different approaches, slightly modified by the Editorial Board.
Solution 1, by Madhav R. Modak.
Let a, b, c be the lengths of the sides BC, CA, AB. Thus b > a > c. Let, with
A as origin, u, v denote the unit vectors along AB and AC respectively. Let
t = (a + b + c)/4. By data, M C + CN = t so that

CN = t − a/2 = (b + c − a)/4.

Similarly, BL = (b + c − a)/4. Also,

AN = b − CN = (a + 3b − c)/4

and
AL = c − BL = (a − b + 3c)/4.
Next,
AK = b − CK = b − (t + CN ) = (b − c)/2.

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 Solutions /95

So the position vectors of points B, C, L, N, K, and M are respectively

a − b + 3c
b = c u, c = b v, l= u,
4
a + 3b − c b−c 1 c b
n= v, k = v, m = (b + c) = u + v.
4 2 2 2 2
Let P be the midpoint of LN, so that its position vector is
1 a − b + 3c a + 3b − c
p= (l + n) = u+ v.
2 8 8
It follows that
a − b + 3c −a + b + c
p−k= (u + v) and m − p = (u + v) .
8 8
Therefore, KP is parallel to P M , i.e. the points K, P, M are collinear, which
achieves the proof. This proves also that the line KM is parallel to the bisector
of the angle ∠BAC of direction vector u + v.

Solution 2, by Bing Jian.

Let P be the midpoint of LN . We need to prove that P , K and M are collinear and
we will leverage barycentric coordinates. In the absolute barycentric coordinate
system relative to the vertices A, B, C, we have
1 1 1
A = (1, 0, 0), B = (0, 1, 0), C = (0, 0, 1) and M = · (B + C) = (0, , ).
2 2 2
Now let’s find the barycentric coordinates of L, N and K.
Let |BC| = a, |AC| = b, and AB| = c with b > a > c. Since points K, L, M , N
divide the perimeter of ABC into 4 equal parts, we have
b+c−a b−c
|BL| = |CN | = , |AK| = ,
4 2
which gives the following barycentric coordinates:
b+c b−c b+c−a 3b + a − c b + c − a 3c + a − b
Å ã Å ã Å ã
K= , 0, ,N= , 0, , L= , ,0 ,
2b 2b 4b 4b 4c 4c
and therefore
1 (b + c − a)(b + c) 3c + a − b 3b + a − c
Å ã
P = · (N + L) = , , .
2 8bc 8c 8b
Then we notice that
b+c 1
Å ã
c
K −M = ,− ,− ,
2b 2 2b
3c + a − b b+c 1 3c + a − b
Å ã
c
K −P = · ,− ,− = · (K − M ),
4c 2b 2 2b 4c

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96/ Solutions

which implies P , K and M are collinear. In fact, we have

b+c−a 3c + a − b
P = ·K + · M.
4c 4c
This completes our proof.

4868. Proposed by Michel Bataille.

Let k ∈ [−1, 1] and let a, b, c be real numbers such that a2 + b2 + c2 = 1. Find the
minimal and maximal values of a3 + b3 + c3 + kabc.
We received 17 submissions, 9 correct and complete. We present the solution by
Marie-Nicole Gras, slightly altered by the editor.
For all k, if a = 1, b = 0, c = 0 then a3 + b3 + c3 + kabc = 1, and if a = −1,
b = 0, c = 0 then a3 + b3 + c3 + kabc = −1. We will prove that if k ∈ [−2, 2] and
a2 + b2 + c2 = 1, then

− 1 ≤ a3 + b3 + c3 + kabc ≤ 1, (1)

which will prove that −1 (respectively 1) is the lower (respectively upper) bound.
It is obvious that if a, b, c, k are nonnegative, then for all u, v, w, t = ±1,

−a3 − b3 − c3 − kabc ≤ ua3 + vb3 + wc3 + tkabc ≤ a3 + b3 + c3 + kabc.

To prove (1), it sufficies to prove that, if a, b, c, k are nonnegative, then

a3 + b3 + c3 + kabc ≤ 1

and since 0 ≤ k ≤ 2, it sufficies to prove that G = a3 + b3 + c3 + 2abc ≤ 1.

By the Cauchy-Schwarz inequality

(a2 + b2 )(a2 + c2 ) ≥ (a2 + bc)2 and b2 + a2 )(b2 + c2 ) ≥ (b2 + ac)2 ,

so that
G = a(a2 + bc) + b(b2 + ac) + c3
» p p √
≤ a (a2 + b2 )(a2 + c2 ) + b b2 + a2 b2 + c2 + c c4
p »
≤ a2 + b2 + c2 (a2 + b2 )(a2 + c2 ) + (b2 + a2 )(b2 + c2 ) + c4
p
= 1 × a4 + b4 + c4 + 2a2 b2 + 2a2 c2 + 2b2 c2
»
= (a2 + b2 + c2 )2
= a2 + b2 + c2 = 1.

Editor’s Comments. Vivek Mehra noted that a generalization of this problem

appeared as Crux Problem 3654 with a solution in Vol. 38, No. 6.

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 Solutions /97

4869. Proposed by Leonard Giugiuc and Mohamed Amine Ben Ajiba.

Let ABC be a non-obtuse triangle with area 1 and side-lengths a, b, c. Let n be a
fixed non-negative real number. Find the minimum value of
2n 1 1 1
+ 2 + 2 + 2.
a2 2
+b +c 2 a b c

There were 4 correct solutions and 3 incomplete solutions. We will present several
approaches.
Solution by UCLan Cyprus Problem Solving Group.
√
When n ≥ 3, the minimum value is n + 1, and when n ≤ 3, the minimum is
1
4 (n + 5).

Since the triangle is acute, the square of each side does not exceed the sum of the
squares of the other two sides. Thus x = 12 (b2 + c2 − a2 ), y = 21 (c2 + a2 − b2 ) and
z = 12 (a2 + b2 − c2 ) are all nonnegative and (a2 , b2 , c2 ) = (y + z, z + x, x + y).
The area of the triangle is given by one quarter of the squareroot of

2(a2 b2 + b2 c2 + c2 a2 ) − (a4 + b4 + c4 ) = xy + yz + zx.

The problem is to find the minimum value of

1 1 1
Å ã
n
P = + + + ,
x+y+z y+z z+x x+y
subject to the conditions that x, y, z > 0 and xy + yz + zx = 4.
Let s = x + y + z and p = xyz. Then
n (x2 + y 2 + z 2 ) + 3(xy + yz + zx)
P = +
s (x + y)(y + z)(z + x)
n (x + y + z)2 + (xy + yz + zx)
= +
s (x + y + z)(xy + yz + zx) − xyz
n s2 + 4 n s2 + 4
= + ≥ +
s 4s − p s 4s
n+1 s √
= + ≥ n + 1.
s 4
The last inequality is due to the arithmetic-geometric means inequality. Equality
holds if and only if xyz = 0 and s2 = 4(n + 1).
√
Suppose, wolog, z = 0. Then c2 = x + y = s = 2 n + 1, a2 = y and b2 = x. Since
also a2√
b2 = xy + yz + zx = 4, a2 and b2 are roots of the quadratic polynomial,
2
t − (2 n + 1)t + 4, whose discriminant is 4(n − 3). Therefore, when n ≥ 3, the
inequality is satisfied when
√ √ »√ √ » √
Å» ã
(a, b, c) = n + 1 + n − 3, n + 1 − n − 3, 2 n + 1 .

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98/ Solutions

When n ≤ 3, this lower bound is not achievable. Let f (s) = (n + 1)/s + s/4. Then

f 0 (s) = −(n + 1)/s2 + 1/4 ≥ −4/s2 + 1/4.

When s ≥ 4, f 0 (s) ≥ 0, so that f (s) ≥ f (4) = (n + 5)/4, so that P ≥ (n + 5)/4.

We show that this is an attainable lower bound when n ≤ 3 and s ≤ 4.
When s ≤ 4, then n/s ≥ n/4 so that it is enough to prove that

s2 + 4 5
≥ ,
4s − p 4
or 4s2 − 20s + 16 + 5p ≥ 0.
By Schur’s inequality, we have, for x, y, z ≥ 0, that

x3 + y 3 + z 3 + 3xyz ≥ x2 (y + z) + y 2 (z + x) + z 2 (x + y),

with equality if and only if (x, y, z) = (t, t, t), (0, t, t), (t, 0, t), (t, t, 0) for some
nonegative real t. Since

x3 + y 3 + z 3 = (x + y + z)3 − 3[x2 (y + z) + y 2 (z + x) + z 2 (x + y)] − 6xyz,

then

s3 − 3p ≥ 4[x2 (y + z) + y 2 (x + z) + z 2 (x + y)]
= 4[(x + y + z)(xy + yz + zx) − 3xyz] = 16s − 12p,

and
16s − s3 s(4 − s)(4 + s)
p≥ = .
9 9
Thus
5s(4 − s)(4 + s)
4s2 − 20s + 16 + 5p ≥ 4(1 − s)(4 − s) +
9
(4 − s)(5s2 − 16s + 36)
=
9
(4 − s)((2s − 4)2 + s2 + 20)
= ≥ 0.
9
Therefore

P = n/s + (s2 + 4)/(4s − p) ≥ n/4 + (s2 + 4)/(4s − p) ≥ (n + 5)/4.

For equality to hold, we require that s = 4, at least two of x, y, z to be nonzero

and equal, and
5 s2 + 4 20
= = .
4 4s − p 16 − p
In turn, this forces p = 0. Wolog, let z = 0 and x = y. Then a2 = b2 and
a2 + b2 = c2 =√x + y = s = 4. Therefore the lower bound of (n + 5)/4 is attained
when a = b = 2 and c = 2.

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 Solutions /99

Summary of the solution by M. Bello, M. Benito, Ó. Ciaurri and E. Fernández,

jointly.
2 2
Without loss of generality, assume that √ a ≥ b ≥ c and let u =2 b + c and
v = bc cos A. Then bc = 2 csc A ∈ [2, 4/ 3], u ≥ 2bc ≥ 4, u − 2v = a > 0, and
√
…
p 4 p
v = bc 1 − sin2 A = bc 1 − 2 2 = b2 c2 − 4 ∈ [0, 2/ 3].
b c
The quantity to be minimized is
n 1 u
P (u, v) = + + 2 .
u − v u − 2v v + 4
Since b2 and c2 are real roots of the quadratic t2 − ut + (v 2 + 4), we must have
u2 − 4v 2 − 16 ≥ 0,
1 p
b2 = (u + u2 − 4v 2 − 16),
2
1 p
c2 = (u − u2 − 4v 2 − 16).
2
√
Since a2 ≥ b2 , u − 4v ≥ u2 − 4v 2 − 16, whence 5v 2 − 2uv + 4 ≥ 0.

√ is to minimize the function P (u, v) subject to the constraints u ≥ 4,

The problem
0 ≤ v ≤ 2/ 3, u2 − 4v 2 − 16 ≥ 0 and 5v 2 − 2uv + 4 ≥ 0.
The equations u2 − 4v 2 − 16 = 0 and 5v 2 − 2uv + 4 = 0, with u, v ≥ 0, describe
branches, H1 and H2 respectively,
√ √ of two hyperbolae in the positive quadrant,
tangent at√the point (8/ 3, 2/
√ 3), with
√ H2 to the right of H1 and tangent to the
line u = 2 5 at the point (2 5, 2/ 5). The constraint region is bounded by the
curves with equations v = 0, 5u2 − 2uv + 4 = 0 and u2 − 4v 2 = 16.
Since the partial derivatives of G(u, v) both vanish only when 5u2 − 2uv + 4 = 0,
the minimizing point must lies on one of the bounding curves. When v = 0,
n+1 u √
P (u, 0) = + ≥ n + 1,
u 4
√
√ only if u = 2 n + 1. Since u ≥ 4, this applies only when n ≥ 3.
with equality
When n < 3, the derivative of P (u, 0) is positive when u > 4, so
P (u, 0) ≥ P (4, 0) = (n + 5)/4.
√ √ √
On the√arc of H1 between (4, 0) and (8/ 3, 2/ 3), we have 4 ≤ u ≤ 8/ 3 and also
v = 21 u2 − 16. The function G(u, v) along this arc is increasing so its minimum
value of (n + 5)/4 is assumed at (4, 0).
The analysis of P (u, v) on the arc of H2 is
√ complicated; it can be shown that on
this arc P (u, v) > (n + 5)/4. When u > 2 5 and (u, v) lies below H2 , we have
p p
0 ≤ [u − u2 − 20]/5 = 4/(u + u2 − 20 < 4/u.
Therefore, subject to the constraint, limu→∞ P (u, v) = ∞. The answer follows.

Copyright © Canadian Mathematical Society, 2024

100/ Solutions

Solution for n ≥ 3, by Theo Koupelis.

Let the triangle be ABC and let AD be the altitude from A to BC. Let x = BD,
y = DC and d = AD. Then a = x + y, ad = 2, b2 = d2 + y 2 , c2 = d2 + x2 and

a2 = b2 + c2 − 2bc cos A ≤ b2 + c2 = 2d2 + x2 + y 2

8
= 2d2 + a2 − 2xy = a2 + 2 − 2xy.
a

It follows that xy ≤ 4/a2 , with equality if and only if A = 90◦ . Since a2 =

(x + y)2 ≥ 4xy, we also have that xy ≤ a2 /4.

2n 1 a2 + a82 − 2xy
P = 8 + + 16
2a2 + a2 − 2xy a2 a4 + a42 (a2 − 2xy) + x2 y 2
(n + 1)a4 + (4 − xya2 ) a6 + 2a2 (4 − xya2 )
= + .
a2 [a4 + (4 − xya2 )] 4a4 + (4 − xya2 )2

Let t = (4 − xya2 )/4. Then 0 ≤ t ≤ 1, and thus

(n + 1)a4 + 4t a6 + 8a2 t (n + 1)a4 + 4t2 a6 + 8a2 t

P = 2 2
+ 4 2
≥ 2 2
+
a (a + 4t) 4(a + 4t ) a (a + 4t) 4(a4 + 4t)
1
= 2 4 · [4(n + 1)a4 + (a4 + 4t)2 ]
4a (a + 4t)
1 » √
≥ 2 4 · 2 · 4(n + 1)a4 · (a4 + 4t) = n + 1.
4a (a + 4t)

Equality occurs when t = 0 or t = 1 and

4(n + 1)a4 = (a4 + 4t)2

or
√
a4 − (2 n + 1)a2 + 4t = 0.
When t = 1, this quadratic has discriminant 4(n − 3), and so the equation is
solvable for n ≥ 3. In this case, we find that xy = 0. Suppose x = 0, then
√ √
a2 = n + 1 ± n − 3;

4 √ √
b2 =2
= n + 1 ∓ n − 3;
a
√
c = d + y 2 = b2 + a2 = 2 n + 1.
2 2

√
When t = 0, then a2 = 2 n + 1 = b2 + c2 , xy = 4/a2 , x2 + y 2 = a2 − (8/a2 ).
Therefore. say,
√ √
2 a4 − 8 ∓ a2 a4 − 16 2 a4 − 8 ± a2 a4 − 16
x = and y = .
2a2 2a2

Crux Mathematicorum, Vol. 50(2), February 2024

 Solutions /101

For the solutions to be real, we require that a2 ≥ 4 and n ≥ 3. In this case

√
a2 = 2 n + 1;
√
a2 a4 − 16 √ √
b2 = ± = n + 1 ± n − 3;
2 √ 2
a 2
a4 − 16 √ √
c2 = ∓ = n + 1 ∓ n − 3,
2 2
essentially as was the situation when t = 0.

4870*. Proposed by Borui Wang.

1
Define the series {an } by the following recursion: a1 = 1, an+1 = an + for
q · an
n > 0, q > 0. Find the constant number c(q) such that
»
lim (an − c(q) · n) = 0.
n→∞

We received 9 submissions and they were all complete and correct. We present the
solution by the majority of solvers.
We have ã2
1 2 1
Å
a2n+1 = an + = + a2n + 2 2 .
q · an q q · an
Then, by induction, it follows that
n−1
2 1 X 1
a2n = 1 + (n − 1) + 2 . (1)
q q a2k
k=1

In particular, for all n ∈ N, we have

»
an > 2(n − 1)/q. (2)

On the other hand, from equation (1) and inequality (2), we deduce that
n−1 n−1
2 1 X 1 2 1 1 X 1
a2n = 1 + (n − 1) + 2 < 1 + (n − 1) + + .
q q a2k q q2 2q n−1
k=1 k=2
Pn 1
It is well-known that Hn := k=1 n ∼ log n. Now it follows from the squeeze
theorem that »
lim (an − 2n/q) = 0.
n→∞

We conclude that c(q) = 2/q.

Copyright © Canadian Mathematical Society, 2024