Volume/tome 50, issue/numéro 7

September/septembre 2024
Crux Mathematicorum is a problem-solving journal at the secondary and university undergraduate levels,
published online by the Canadian Mathematical Society. Its aim is primarily educational; it is not a research
journal. Online submission:
https://publications.cms.math.ca/cruxbox/

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cycle universitaire publiée par la Société mathématique du Canada. Principalement de nature éducative,
le Crux n’est pas une revue scientifique. Soumission en ligne:
https://publications.cms.math.ca/cruxbox/

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ISSN 1496-4309 (électronique)

Supported by / Soutenu par :

• Intact Financial Corporation
• University of the Fraser Valley

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
Almaz Butaev University of the Fraser Valley
Mateusz Buczek Warsaw, Poland
Ana Duff Ontario Tech University
Andrew McEachern York University
Chi Hoi Yip Georgia Institute of Technology
Matt Olechnowicz Concordia University
Vasile Radu Birchmount Park Collegiate Institute
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

339 Editorial Kseniya Garaschuk

340 MathemAttic: No. 58
342 Problems: MA281–MA285
342 Solutions: MA256–MA260
348 Problem Solving Vignettes: No. 33 Shawn Godin
353 The Third Marble Revisited
Geoffrey W. Brown and Adam C. Brown
356 Olympiad Corner: No. 425
356 Problems: OC691–OC695
358 Solutions: OC666–OC670
363 The Sneakiest Ugly Integral Trick Yuepeng Alex Yang
367 Problems: 4961–4970
371 Solutions: 4911–4920

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
 Editorial /339

EDITORIAL
We would like to dedicate this issue to Edward Wang, who is stepping away from
Crux after years of service.
Searching through the issues, it seems Edward’s first appearance in Crux was as a
proposer of problem 1019 in February 1985. Edward joined the Editorial Board in
1993, staying with the publication for over 30 years. But Crux wasn’t Edward’s
only contribution to the Canadian problem solving community as many remember
him being quite involved in many initiatives, such as the Canadian International
Math Olympiad (IMO) team: he was on the Canadian Team in 2001 as Observer
and in 2004 as Deputy, he has organized at least one Training Camp at Wilfrid
Laurier University and has always been around to help. Various stories I heard
over the last month connect him to many mathematical events; below are just a
couple of them passed down to me.
There was one very particular way in which Edward made a contribution to the
IMO in 1998. One of the members selected for the team was Jessie Lei, a student
of Vincent Massey Collegiate in Windsor, Ontario. The IMO was to be held in
Taipei, Taiwan that year, so the entire Canadian delegation needed visas. However,
Taiwan refused to issue a visa for Jessie because while she was a permanent resident
in Canada (which made her eligible), her citizenship was the People’s Republic of
China. An approach to the Taiwanese consulate in Toronto turned out to be
fruitless. When Edward found out, it turned out that not only was he from
Taiwan, but that his mother was or had been a Senator in the Parliament. So he
got in touch with his mom and the visa came through.
The year was 2000 or 2003, when Andy Liu was on the IMO team and Edward
organized the training camp at Wilfrid Laurier University. One day at the training
camp, an informal table tennis tournament was held among the six students on
the team, plus Edward and Andy. Despite the students’ big age advantage, the
two players in the final match were Ed and Andy. Who prevailed? That shall
remain one of history’s mysteries.
When I myself started with Crux over 10 years ago, I inherited a knowledgeable
and active Editorial Board, very few of whose members I met in person. We quickly
developed relationships through email and ongoing work. Edward has always been
one of my favourite people to correspond with. He would tell me that it was hard
to keep up with technology in his advanced age, yet he smoothly adjusted to all
the new tools introduced to Crux, easily navigating Dropbox and Google sheets.
On top of math, we bonded over random facts of life, such as both of Edward and
my daughter being born in the year of the Monkey in Chinese Zodiac. We would
often exchange proverbs from our native cultures and languages, and it’s always
fun explaining to people how I know Chinese proverbs that I occasionally use.
Thank you Edward for your contributions, advice and camaraderie over the years.
Kseniya Garaschuk

Copyright © Canadian Mathematical Society, 2024

340/ MathemAttic

MATHEMATTIC
No. 58
The problems featured 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 November 15, 2024.

MA281. Proposed by Trinh Quoc Khanh.

Given a triangle ABC, let E, F be the feet of altitudes from B, C. The circle
centered at B with radius BE intersects segments BA, BC at X, Z. The circle
centered at C with radius CF intersects segments CA, CB at Y, T . XZ intersects
Y T at K. Let I, J be the incenters of 4XEF, 4Y EF , respectively. Prove that
AK is perpendicular to IJ.

MA282. Proposed by Neculai Stanciu.

Prove that 22n+5 + 9n2 + 3n + 4 is divisible by 18 for any non-negative integer n.

MA283. The integers from 1 to 9 are listed on a blackboard. If an additional

m eights and k nines are added to the list, the average of all of the numbers in the
list is 7.3. Find the value of k + m.

MA284. The lengths of all six edges of a tetrahedron are integers. The
lengths of five of the edges are 14, 20, 40, 52, and 70. How many possible lengths
for the sixth edge are there?

MA285. In parallelogram ABCD, AB = a and BC = b, a > b. Angles

A, B, C, D are bisected. The intersection points of these angle bisectors are the
vertices of quadrilateral P QRS. Prove that P R = a − b.

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

Crux Mathematicorum, Vol. 50(7), September 2024

 MathemAttic /341

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 novembre 2024.

MA281. Soumis par Trinh Quoc Khanh.

Étant donné un triangle ABC, notons E et F les pieds des hauteurs à partir de
B et C, respectivement. Le cercle centré en B de rayon BE coupe les segments
BA et BC en X et Z, respectivement. Le cercle centré en C de rayon CF coupe
les segments CA et CB en Y et T , respectivement. De plus, XZ coupe Y T en K.
Soient I et J les centres de 4XEF et 4Y EF , respectivement. Montrez que AK
est perpendiculaire à IJ.

MA282. Soumis par Neculai Stanciu.

Montrez que 22n+5 + 9n2 + 3n + 4 est divisible par 18 pour pour tout entier non-
négatif n.

MA283. Les nombres entiers de 1 à 9 sont inscrits sur un tableau. Si l’on

ajoute m huit et k neuf à la liste, la moyenne de tous les nombres de la liste est
7, 3. Trouvez la valeur de k + m.

MA284. Les longueurs des six arêtes d’un tétraèdre sont des nombres entiers.
Les longueurs de cinq des arêtes sont respectivement 14, 20, 40, 52 et 70. Combien
y a-t-il de longueurs possibles pour la sixième arête ?

MA285. Dans le parallélogramme ABCD, AB = a et BC = b, a > b. Les

angles A, B, C et D sont bissectés. Les points d’intersection de ces bissectrices
sont les sommets du quadrilatère P QRS. Montrez que P R = a − b.

Copyright © Canadian Mathematical Society, 2024

342/ MathemAttic

MATHEMATTIC
SOLUTIONS
Statements of the problems in this section originally appear in 2024: 50(3), p. 59–61.

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

the rightmost decimal digit into the leftmost position. Find the smallest such
number.
Originally question 10 from the 2011 Manitoba Mathematical Contest.
We received 3 submissions, all of which were correct and mostly complete. We
present the solution by Soham Das, adapted for clarity and completeness.
Let xn−1 xn−2 . . . x1 x0 be the number we are seeking. According to the problem
| {z }
n digits
statement,

x0 × 10n−1 +xn−1 × 10n−2 + xn−3 × 10n−3 + · · · + x1

= 4 xn−1 × 10n−1 + x2 × 10n−2 + · · · + x1 × 10 + x0

Therefore,

x0 10n−1 − 4 = 39 xn−1 × 10n−2 + x2 × 10n−3 + · · · + x1

x0 10n−1 − 4


⇒ = xn−1 × 10n−2 + x2 × 10n−3 + · · · + x1
39
Since the right side of the equation is an integer, x0 (10n−1 − 4) is divisible by
39 = 3 × 13. Since x0 is a single-digit number, it is relatively prime with 13, and
so 10n−1 − 4 is divisible by 13. By trial end error we determine that the minimum
such value of n − 1 is 5, i.e., n = 6. Therefore, the number sought is of the form
x5 x4 x3 x2 x1 x0 , where x5 6= 0, and its quadruple is x0 x5 x4 x3 x2 x1 . Hence,

x0 105 − 4


= x5 × 106−2 + x4 × 106−3 + x3 × 106−4 + x2 × 106−5 + x1 × 106−6
39
⇒ 2564x0 = x5 × 10000 + x4 × 1000 + x3 × 100 + x2 × 10 + x1
Since x5 6= 0 and 2564x0 < 10000 for x ≤ 3, the smallest possible value for x0 is
4. Solving for that potential value of x0 in the last equation, we get

2564 × 4 = 1000x5 + 1000x4 + 100x3 + 10x2 + x1

and so x5 = 1, x4 = 0, x3 = 2, x2 = 5, x1 = 6. Therefore, the number sought is

x5 x4 x3 x2 x1 x0 = 102564.

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 MathemAttic /343

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 .
Originally question 10 from the 2012 Manitoba Mathematical Contest.
We received 2 correct submissions, out of which we present the one by Luyu Han,
slightly expanded by the editor.
Since the three rectangles are geometrically similar but not congruent, exactly one
has one side length equal to one and we are in the situation as in the following
figure:

a A

1−a B C

b 1−b

Since the three rectangles are similar, we have

1 b 1−a
= = .
a 1−a 1−b
It follows that 1 − a = ab and 1 − b = a(1 − a) and so

b = 1 − a(1 − a) = a2 + 1 − a.

Using the fact that 1 − a = ab, we get b = a2 + ab and

a2 = b − ab = b(1 − a) = ab2

which implies a = b2 .
It follows from 1 − b = a(1 − a) and 1 − a = ab that

AC = a(1 − a)(1 − b) = a2 (1 − a)2 = a4 b2 .

Since a = b2 , we have AC = a5 . On the other hand, since 1 − a = ab and a = b2 ,

we have
B = b(1 − a) = b(ab) = ab2 = a2 .
Then (AC)2 = (a5 )2 = a10 and B 5 = (a2 )5 = a10 . Thus (AC)2 = B 5 .

Copyright © Canadian Mathematical Society, 2024

344/ MathemAttic

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.

Originally question 10 from the 2013 Manitoba Mathematical Contest.

We received 5 submissions of which 4 were correct and complete. We present the

solution by Soham Das.

Denote by O the origin (0, 0) and by B(−1, 0) and C(3, 0) the centres of the circle
of radius 3 and the circle of radius 1 respectively. Denote by Γ the circle which is
tangent to the given circles (Γ is shown in a dashed red line), by A the centre of
Γ and by E, F and G the points of tangency as in the diagram below. Let r be
the radius of Γ.

Using the properties of tangency we have

BA = BE + EA = 3 + r,
CA = CF + F A = 1 + r, and
OA = OG − GA = 4 − r.

By Stewart’s Theorem for the length of a cevian

AB 2 · OC + AC 2 · BO
AO2 = − BO · OC.
BO + OC

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 MathemAttic /345

Substituting the known lengths we get

(3 + r)2 · 3 + (1 + r)2 · 1
(4 − r)2 = − 1 · 3,
1+3
which we multiply out to solve for r:

4(4 − r)2 = 3(3 + r)2 + (1 + r)2 − 12 ⇔

4(16 − 8r + r2 ) = 3(9 + 6r + r2 ) + (1 + 2r + r2 ) − 12 ⇔
64 − 32r + 4r2 = 28 + 20r + 4r2 − 12 ⇔
48 = 52r ⇔
12
r= .
13
12
Therefore the radius of the circle Γ tangent to the three given circles is 13 .

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.
Originally question 10 from the 2014 Manitoba Mathematical Contest.
We received 5 submissions, 4 of which were correct. We present the solution by
Corneliu Manescu-Avram for part a) and Mihika Bansal for part b).
a) Working modulo 12, we have

7a + 12b = 365 ⇒ 7a ≡ 5 mod 12 ⇒ a ≡ 7 · 7a ≡ 7 · 5 ≡ −1 mod 12

hence a = 12t − 1, with t a positive integer. Then

365 − 7a 365 − 7(12t − 1)

b= = = 31 − 7t
12 12
From b ≥ 0, we deduce that t ≤ 4, and therefore the set of solutions to the equation
is
S = {(11, 24), (23, 17), (35, 10), (47, 3)}

b) Apply the Frobenius 2-coin problem solution (Chicken McNugget Theorem):

For any two relatively prime integers m and n, the largest integer that cannot be
expressed as am + bn for some integers a, b ≥ 0 is mn − m − n.

Copyright © Canadian Mathematical Society, 2024

346/ MathemAttic

Since 7 and 12 are relatively prime,

mn − m − n = (7)(12) − 7 − 12 = 65

so 65 is the largest integer that is not of the form 7a+12b for some integers a, b ≥ 0.
Therefore the largest c is 65.

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.

Originally question 10 from the 2015 Manitoba Mathematical Contest.

We received 4 solutions, 3.5 of which were completely correct. We present the

solutions of Samya Chakrabarti (part a) and Mihika Bansal (part b).

a) S = 2! 3! 4! 5! 6! 7! 8! 9! 10! 11! 12!

prime
2! 3! 4! 5! 6! 7! 8! 9! 10! 11! 12!
factors
2 1 1 3 3 4 4 7 7 8 8 10
3 0 1 1 1 2 2 2 4 4 4 5
5 0 0 0 1 1 1 1 1 2 2 2
7 0 0 0 0 0 1 1 1 1 1 1
11 0 0 0 0 0 0 0 0 0 1 1

S = 256 × 326 × 511 × 76 × 112

= (252 × 324 × 510 × 76 × 112 ) × (24 × 32 × 5)
= (226 × 312 × 55 × 73 × 11)2 × 6!

Hence, N = 6.

b) add 1! to make sets of 2

1! 2! 3! 4! 5! 6! 7! 8! · · · (4n − 1)! (4n)!

= (1! 2!)(3! 4!)(5! 6!)(7! 8!) · · · ((4n − 1)! (4n)!)
= (1! 1! 2)(3! 3! 4)(5! 5! 6)(7! 7! 8) · · · ((4n − 1)! (4n − 1)! (4n))

Crux Mathematicorum, Vol. 50(7), September 2024

 MathemAttic /347

2=2·○
1 4=2·○
2 6=2·○
3 8=2·○
4 4n = 2 · 2n

= (1!)2 (3!)2 (5!)2 (7!)2 · · · ((4n − 1)!)2 · 22n · (2n)!

| {z } | {z }
square factorial

N = 2n
M = 1! 3! 5! · · · (4n − 1)! · 2n

Editor’s Comments. The representation of the product of factorials in the form

M 2 N ! is generally not unique. Also, it is not just numbers of the form 4n for
which the product of factorials can be written this way. For example,

1! 2! 3! 4! 5! 6! 7! 8! 9! 10! 11! 12! 13! 14!

= 13092485195995938186854400000002 × 8!
= 4364161731998646062284800000002 × 9!

By the way, the expression 2 · 4 · 6 · 8 · · · (4n) = 22n · (2n)! appearing in the solution
to part b) is called a “double factorial”. The double factorial k!! of a number k is
the product of all the numbers between 1 and k that have the same parity as k
(even or odd). For example, 5!! = 1 · 3 · 5 and 6!! = 2 · 4 · 6.

Copyright © Canadian Mathematical Society, 2024

348/ Problem Solving Vignettes

PROBLEM SOLVING
VIGNETTES
No. 33
Shawn Godin
As Easy As Falling Off of a . . .

Logarithms are a topic that Canadian students usually face near the end of their
high school journey. Problems involving logarithms will sometimes sneak their way
into upper level high school mathematics contests. I was inspired for this column
by a problem that I was marking for the CEMC Summer Problem Solving Course
this year. The course is based on the free Problem Solving and Mathematical
Discovery course that appears on the CEMC courseware site, which would be
of interest to readers of this column. The particular problem got me thinking
because there are usually several ways to attack these types of problems, which
is nice. There are also some useful properties of logarithms that are not usually
taught in high school that follow easily from that which is taught. Let’s start with
a quick review of the basic definition of logarithms.

Given a, b, x ∈ R with b > 0, b 6= 1 and ba = x, we can define the logarithm with

base b of x by logb (x) = a. That is, the logarithm of a number is the exponent that
the base must be raised to in order to yield the original number. It is not hard
to see that the functions f (x) = bx and g(x) = logb (x) are inverses of each other.
Let’s jump right into a problem. This problem, which I mentioned earlier, was
from the second assignment of the 2024 CEMC Summer Problem Solving Course.
Ä √ ä Ä √ ä
5. Determine all real x that satisfy log2x 48 3 3 = log3x 162 3 2 .

A common technique for dealing with equations involving logarithms is to rewrite

the problem in exponential form. Conversely, sometimes when a problem involves
lots of powers, looking at things in logarithmic form can be useful. For the problem
at hand, if we let
Ä √3
ä Ä √
3
ä
log2x 48 3 = log3x 162 2 = k

then we can rewrite the two sides of the equation in exponential form as

√
(2x)k = 48 3,
3
(1)
√
(3x)k = 162 2.
3
(2)

Crux Mathematicorum, Vol. 50(7), September 2024

 Shawn Godin /349

Dividing these two equations yields

√
(2x)k 48 3 3
= √
(3x)k 162 3 2
Å ãk …
2x 8 3 3
=
3x 27 2
Å ãk Å ã 83
2 2
=
3 3
from which we can conclude that k = 83 . Substituting this into (1) yields
8 √
3
(2x) 3 = 48 3
8 8 4
2 3 x 3 = 24 × 3 3
8 4 4
x3 = 23 × 33
8 4
x 3 = (2 × 3) 3
√
x= 6
This can be verified by noting, for the left hand side, that
√ 3 1
2x = 2 6 = 2 2 × 3 2
and
√ 4
Ä √ ä 83
48 3 = 24 × 3 3 = 2 6
3

so Ä √ ä 8
3
log2x 48 3 = .
3
8
We can similarly show that the right hand side is also 3. I will leave that as an
exercise to interested readers.
When you first encounter logarithms in school, we are also shown the laws of
logarithms which are just the power laws that we encounter earlier in our education
rewritten in logarithmic form. Thus we get:
bx × by = bx+y ⇔ logb (X × Y ) = logb (X) + logb (Y )
x y x−y
b ÷b =b ⇔ logb (X ÷ Y ) = logb (X) − logb (Y )
x y x×y
(b ) = b ⇔ logb (X Y ) = Y logb (X)

for b, x, y, X, Y ∈ R with b, X, Y > 0 and b 6= 1. You may want to reexamine the

first problem to see if any of these laws would be helpful. Let’s look at another
problem, this one is from the 2023 Euclid contest.

8. (b) Determine all real values of x for which

» … x
log2 (x) · log2 (4x) + 1 + log2 (x) · log2 +9=4
64

Copyright © Canadian Mathematical Society, 2024

350/ Problem Solving Vignettes

Using the laws of logarithms, we can rewrite this as

» »
log2 (x) · (log2 (4) + log2 (x)) + 1 + log2 (x) · (log2 (x) − log2 (64)) + 9 = 4

If we let y = log2 (x), and note that 4 = 22 and 64 = 26 , we can rewrite the
equation as
» »
y(2 + y) + 1 + y(y − 6) + 9 = 4
p p
y 2 + 2y + 1 + y 2 − 6y + 9 = 4
» »
(y + 1)2 + (y − 3)2 = 4
|y + 1| + |y − 3| = 4

We need to look at three cases:

Case 1: y ≥ 3.
The equation becomes

(y + 1) + (y − 3) = 4
2y − 2 = 4
y=3

so log2 (x) = 3 yielding x = 8.

Case 2: y ≤ −1.
The equation becomes

−(y + 1) − (y − 3) = 4
−2y + 2 = 4
y = −1

so log2 (x) = −1 yielding x = 21 .

Case 3: −1 < y < 3.

The equation becomes

(y + 1) − (y − 3) = 4
4=4

so all values in this range are solutions.

Putting the three cases together we see that the solutions to the equation are
1
2 ≤ x ≤ 8.

In the first problem you might have noticed that the two logarithms had different
bases. If the bases were the same, we could have taken the inverse of both sides,

Crux Mathematicorum, Vol. 50(7), September 2024

 Shawn Godin /351

eliminating the logarithms, and making the problem much easier. Since the bases
were different, the two logarithms had different inverses. It would have been useful
to have the logarithms with the same base, so let’s try to take a logarithm and
write it with a different base.
Suppose we are looking at logb (a), and want to write it in terms of a logarithm
base c. Letting x represent the expression we get

logb (a) = x

which we can write in exponential form (by raising b to the exponent of each side
of the equation) as
a = bx .
Taking logarithms with base c of both sides yields

logc (a) = logc (bx ) = x logc (b) = logb (a) · logc (b)

which can be rewritten as

logc (a)
logb (a) =
logc (b)
which is called the change of base formula for logarithms. It is usually helpful to
change the base to 10 so we can use common logarithms. Writing log(x) without
the base usually means the base is 10. In some cases it may mean that the base
is e, but the document in which it appears will usually indicate that if that is
the case. In most cases when the base is e = 2.718281 . . . , the natural base of
logarithms, we will write ln(x) for the natural logarithm of x.
Let’s use this in a problem. This is from the great book Five Hundred Mathematical
Challenges [1]. The book was authored by current Crux editor and contributor Ed
Barbeau; original editor of the Olympiad Corner and contributor Murray Klamkin;
and contributor William Moser.

Problem 294. Find (log3 (169)) × (log13 (243)) without use of tables.

Utilizing our change of base formula, we get

log(169) log(243)
(log3 (169)) × (log13 (243)) = ×
log(3) log(13)
log(13 ) log(35 )
2
= ×
log(3) log(13)
2 log(13) 5 log(3)
= ×
log(3) log(13)
= 10

A keen eyed reader may have noticed another useful identity hiding within the
solution. That is, for x, y ∈ R with x, y > 0 and x, y 6= 1, then

logx (y) × logy (x) = 1

Copyright © Canadian Mathematical Society, 2024

352/ Problem Solving Vignettes

I will leave it to the reader to supply the proof.

I hope you have enjoyed our look at logarithms. You may want to reexamine the
first problem to see if any of the properties we discussed in the article lead to other
solutions. I leave you with a few problems for your entertainment.

1. Determine all real x that satisfy log2x (80) = log5x (500). [2022 CEMC Prob-
lem Solving Course, Assignment 1, problem 4]
p
2. Which of the following is the value
p of log2 6 + log3 6?
(A) 1p p (B) plog5 6 p (C) 2
(D) log2 3 + log3 2 (E) log2 6 + log3 6
[2020 AMC12B Contest, problem 13]
3. Prove that log10 2 is irrational. [Problem 452 from [1]]
4. There is a unique positive real number x such that the three numbers
log8 (2x), log4 x, and log2 x, in that order, form a geometric progression with
positive common ratio. The number x can be written as m n , where m and n
are relatively prime positive integers. Find m + n.
[2020 AIME I, problem 2]
5. Determine all real numbers a, b and c for which the graph of the function
y = loga (x + b) + c passes through the points P (3, 5), Q(5, 4) and R(11, 3).
[2022 Euclid Contest, problem 6b]
6. Real numbers x and y with x, y > 1 satisfy logx (y x ) = logy x4y = 10.


What is the value of xy? [2024 AIME I, problem 2]
7. Determine all triples (x, y, z) of real numbers that are solutions to the fol-
lowing system of equations: [2024 Euclid Contest, problem 8b]

log9 x + log9 y + log3 z = 2

log16 x + log4 y + log16 z = 1
log5 x + log25 y + log25 z = 0
8. Let k, m, and n be positive integers with the property: for some number x 6=
1, the numbers logk x, logm x, logn x are consecutive terms of an arithmetic
progression. Show that
n2 = (kn)logk m .
[Problem 354 from [1]]

References
[1] Edward J. Barbeau, Murray S. Klamkin, William O.J. Moser, Five Hundred
Mathematical Challenges, The Mathematical Association of America, Washington,
1995.

Crux Mathematicorum, Vol. 50(7), September 2024

 Geoffrey W. Brown and Adam C. Brown /353

The Third Marble Revisited

Geoffrey W. Brown and Adam C. Brown
The motivation for this paper comes from an example in [1] Mathematical Statistics
with Applications, p. 275. The set-up from this example is as follows:
Suppose that an urn contains r red marbles and (N − r) black marbles.
We draw, one at a time and without replacement, a sequence of k mar-
bles from the urn, where k ≤ N . We then ask the following question:
What is the probability that the jth marble drawn is red, and the kth
marble drawn is also red, where j < k ≤ N ?
Solution. The first step of the solution is to define the following random variables.
We let

ß
1 if the ith marble drawn is red
Xi =
0 otherwise

Before tackling the general case, we do two simple examples.

Example 1. Find P (X1 = 1, X2 = 1).
Solution. This case is just drawing two red marbles in succession from the urn,
which clearly has probability

r r−1
· .
N N −1

Example 2. Find P (X2 = 1, X3 = 1).

Solution. We can enumerate this case quite simply, since the first marble drawn
must be either red or black. Thus we see that:

P (X2 = 1, X3 = 1)
= P (X1 = 1, X2 = 1, X3 = 1) + P (X1 = 0, X2 = 1, X3 = 1)
r r−1 r−2 N −r r r−1
= · · + · ·
N N −1 N −2 N N −1 N −2
r(r − 1)
= [r − 2 + N − r]
N (N − 1)(N − 2)
r(r − 1)(N − 2)
=
N (N − 1)(N − 2)
r r−1
= · .
N N −1

Copyright © Canadian Mathematical Society, 2024

354/ The Third Marble Revisited

Example 2 produces the very same answer as we found in Example 1, and it leads
us to conjecture that, for any j < k ≤ N , we have

r r−1
P (Xj = 1, Xk = 1) = · . (1)
N N −1

Now how can we prove our conjecture?

The basic idea to solve the problem is not to stop the drawing at the kth marble,
but rather to draw out all N marbles from the urn. This idea was considered in
[2] The Third Marble, and there the authors showed that all sequences drawn of
length N were equally likely. Thus we restate the following theorem from [2]:
Theorem. The probability of drawing any sequence of r red marbles and N − r
black marbles is the same, and is therefore equal to N1 .
(r)
Thus our sample space consists of Nr equally likely points (or sequences), so we


may now compute P (Xj = 1, Xk = 1) by counting the number of sequences (of r
red and (N − r) black marbles) having an r in the jth and kth positions,

and then
just dividing this count by the total number of sequences, which is Nr .
Now the number of sequences of r red and (N − r) black marbles having an r
in the jth and kth positions is just the number of permutations of r − 2 red and
−2
N − r black marbles. This number is just given by Nr−2


. So finally, we compute
P (Xj = 1, Xk = 1) as the following ratio:
N −2


r−2 (N − 2)! r!(N − r)!
P (Xj = 1, Xk = 1) = N
= ·
r
(r − 2)!(N − r)! N!
r r−1
= · . (2)
N N −1
This is precisely the probability that we conjectured in (1), and so (2) illustrates
the power of working with the full sample space, where each sequence of length N
is equally likely.
Clearly this approach will generalize to t red marbles, where t ≤ r. Thus if we
seek the probability P (Xj1 = 1, Xj2 = 1, . . . , Xjt = 1), we just count the number
of sequences of r red and (N − r) black marbles, where an r occurs in positions
−t
j1 , j2 , . . . , jt . This number is given by Nr−t


, and we again divide this by the total
number of sequences, which is Nr . Thus we see that

N −t


r−t (N − t)! r!(N − r)!
P (Xj1 = 1, Xj2 = 1, . . . , Xjt = 1) = N
= ·
r
(r − t)!(N − r)! N!
r r−1 r − (t − 1)
= · ··· . (3)
N N −1 N − (t − 1)
Now the final probability found on the right hand side of (3) is easily seen to
be P (X1 = 1, X2 = 1, . . . , Xt = 1), which we could calculate directly. Thus (3)

Crux Mathematicorum, Vol. 50(7), September 2024

 Geoffrey W. Brown and Adam C. Brown /355

says that the result holds for any positions of the t red marbles, where t ≤ r and
1 ≤ j1 < j2 < · · · < jt ≤ N . Thus (3) is a very intuitive and appealing result, and
it completes the main body of the paper.

Additional Exercises
1. An urn contains 40 red marbles and 60 black marbles. 23 marbles are drawn
from the urn in sequence. Find the probability that:
(a) The 20th marble drawn is red and the 23rd marble drawn is red.
(b) The 20th marble drawn is red and the 23rd marble drawn is black.
(c) The 20th marble drawn is black and the 23rd marble drawn is red.
(d) The 20th marble drawn is black and the 23rd marble drawn is black.
(e) What do your answers in (a), (b), (c), and (d) add up to?
2. An urn contains 5 red marbles, 5 black marbles, and 1 blue marble. All the
marbles are drawn in sequence from the urn. Find the probability that the blue
marble is drawn between a red and a black marble, in either order.

Answers and Solutions

40 39 40 60 60 40 60 59
Answers to 1. (a) 100 · 99 ; (b) 100 · 99 ; (c) 100 · 99 ; (d) 100 · 99 ; (e) 1.

Solution to 2. The probability of getting the subsequence red, blue, black is

5 1 5
· · ,
11 10 9
and similarly the probability of getting the subsequence black, blue, red is also
given by
5 1 5
· · .
11 10 9
This is true for all 9 (of the 11) possible positions of the blue marble, so the total
probability is given by
5 1 5 5
· · ×2×9= ,
11 10 9 11
the desired probability.

References
[1] Mathematical Statistics with Applications, 7th Edition, Dennis D. Wackerly,
William Mendenhall III, Richard L. Scheaffer, Brooks/Cole Cengage Learn-
ing, 2008.
[2] The Third Marble, G. W. Brown, A. C. Brown, Crux Mathematicorum, Vol.
49, Issue 5, May 2023, pp 243-245.

Copyright © Canadian Mathematical Society, 2024

356/ OLYMPIAD CORNER

OLYMPIAD CORNER
No. 425
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 November 15, 2024.

OC691. Prove that a number written only by zeros and ones, with the
number of ones being at least two, cannot be a perfect square.

OC692. Prove that for any positive integer n, the number

n(2n + 1)(3n + 1) · . . . · (1966n + 1)

is divisible by every prime number less than 1966.

OC693. On the side AC of triangle ABC, point E is chosen. Bisector AL

intersects segment BE at point X. It turns out that AX = XE and AL = BX.
What is the ratio of angles A and B of the triangle?

OC694. Inside the parallelogram ABCD, a point E is marked, lying on the

bisector of angle A, and a point F is marked, lying on the bisector of angle C. It
is known that the midpoint of the segment BF lies on the segment AE. Prove
that the midpoint of the segment DE lies on the line CF .

OC695. Let us call two numbers almost equal if they are equal or differ from
each other by at most 1. A checkered rectangle with side lengths equal to natural
numbers a and b is such that it is impossible to cut out a rectangle along the grid
lines whose area is almost equal to half the area of the original rectangle. What
is the smallest value that the number |a − b| can take?

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

Crux Mathematicorum, Vol. 50(7), September 2024

 OLYMPIAD CORNER /357

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 novembre 2024.

OC691. Montrez qu’un nombre s’écrivant uniquement avec des zéros et des
uns et dont le nombre de uns est au moins égal à deux ne peut pas être un carré
parfait.

OC692. Montrez que pour tout entier positif n, le nombre

n(2n + 1)(3n + 1) · . . . · (1966n + 1)

est divisible par tous les nombres premiers inférieurs à 1966.

OC693. On choisit un point E sur le côté AC du triangle ABC. La médiatrice

AL rencontre le segment BE en un point X. Il s’avère que AX = XE et aussi
AL = BX. Quel est le rapport des angles A et B du triangle ?

OC694. À l’intérieur du parallélogramme ABCD, un point E est inscrit sur

la bissectrice de l’angle A et un point F est inscrit sur la bissectrice de l’angle C.
On sait que le point milieu du segment BF est situé sur le segment AE. Prouvez
que le point milieu du segment DE est situé sur la droite CF .

OC695. On dira de deux nombres qu’ils sont presque égaux s’ils sont égaux ou
s’ils diffèrent l’un de l’autre d’au plus 1. Un rectangle quadrillé dont les longueurs
des côtés sont égales aux entiers naturels a et b est tel qu’il est impossible de
découper un rectangle le long des lignes du quadrillage dont l’aire est presque
égale à la moitié de l’aire du rectangle d’origine. Quelle est la plus petite valeur
que peut prendre le nombre |a − b| ?

Copyright © Canadian Mathematical Society, 2024

358/ OLYMPIAD CORNER

OLYMPIAD CORNER
SOLUTIONS
Statements of the problems in this section originally appear in 2024: 50(2), p. 69–70.

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?
Originally from the Fall 2023, Tournament of Towns, Junior O-Level.
We received 3 correct solutions and we present two of them.
Solution 1, by UCLan Cyprus Problem Solving Group.
No. It could be the case that Clarissa visited 1, 3, 4, 2, 5, 7, 6, 8, 9, 10 in this order
and Marissa visited 1, 3, 2, 4, 7, 5, 6, 8, 9, 10 in this order.
Solution 2, by Roy Barbara.
Let i → j denote the jump from square i to square j. Here is a counterexample:
Jumps of Clarissa: 1 → 4 → 2 → 6 → 7 → 3 → 5 → 8 → 9 → 10.
Jumps of Marissa: 1 → 4 → 6 → 2 → 3 → 7 → 5 → 8 → 9 → 10.
In both cases, the jumps have respective lengths: (3, 2, 4, 1, 4, 2, 3, 1, 1).
Remark. The answer is clearly negative for a 1 × m board, where m ≥ 8 is an
arbitrary integer.

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

over, ∠DAC = ∠ABD and ∠CAB = ∠DBC. Is ABCD necessarily a square?
Originally from the Fall 2023, Tournament of Towns, Junior O-Level.
We received 9 correct solutions and we present 2 of them.
Solution 1, by Oliver Geupel.
The answer is no.
Consider a triangle ABC with angles ∠CAB = 50◦ and ∠ABC = 100◦ . Let point
D be the reflection of point C in the perpendicular bisector of the line segment
AB. Then, the quadrilateral ABCD is an isosceles trapezoid.

Crux Mathematicorum, Vol. 50(7), September 2024

 OLYMPIAD CORNER /359

Hence,
∠DAC = ∠DAB − ∠CAB = ∠ABC − ∠CAB = 50◦ .
By symmetry,

∠ABD = ∠CAB = 50◦ = ∠DAC = ∠DBC.

Consequently, ABCD is a non-square convex quadrilateral with the given proper-

ties.

Solution 2, by UCLan Cyprus Problem Solving Group.

No, it’s not necessary. For example let ABX be an equilateral triangle and let
C, D be the midpoints of XB and XA respectively. Then ABCD is convex, and
CD is parallel to AB. Furthermore, AC and BD are medians and therefore angle
bisectors of the triangle ABC and therefore

∠DAC = ∠ABD = ∠CAB = ∠DBC = 30◦ .

There are of course other examples. In all of them ABCD is an isosceles trapezoid
with AC and BD being bisectors of angles ∠DAB and ∠CBA respectively.

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 ?
Originally from the Fall 2023, Tournament of Towns, Senior O-Level.
We received 7 solutions. We present the solution by Oliver Geupel.
The answer is 3100 . For any positive integer n, let An denote the set of n–digit
positive integers with digits in the set D = {2, 3, 4, 5, 6, 7}, that are divisible by
2n but not divisible by 2n+1 , let Bn denote the set of n–digit positive integers
with digits in D that are divisible by 2n+1 , and let Cn = An ∪ Bn . Clearly,

Copyright © Canadian Mathematical Society, 2024

360/ OLYMPIAD CORNER

C1 = {2, 4, 6}. Let n > 1, and let m be any n-digit positive integer with digits in
D. Write m in the form m = 10n−1 d + r with d ∈ D and an (n − 1)-digit positive
integer r. Then m ∈ Cn holds if and only if either d is odd and r ∈ An−1 , or d is
even and r ∈ Bn−1 . There are three odd digits 3, 5, 7 and three even digits 2, 4,
6. Thus,

|C1 | = 3; |Cn | = 3 (|An−1 | + |Bn−1 |) = 3 |Cn−1 | (n > 1).

We conclude |Cn | = 3n . Hence the result.

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 1 2 −1 0 1
CAC = or CAC = .
−3 −1 −5 0

Originally from the 45th Brazilian Olympiad of Mathematics (2023), University

Level Exam, Problem 4.
We received 4 correct solutions. We present the solution by UCLan Cyprus Prob-
lem Solving Group.
Let us call two matrices X, Y ∈ M2 (Z) integrally similar if there is an invertible
matrix C ∈ M2 (Z) with C −1 ∈ M2 (Z) and CXC −1 = Y .
It is easily seen that integral similarity is an equivalence relation and it is enough
to show that A is integrally similar to
Å ã Å ã
1 2 0 1
A1 = or A2 = .
−3 −1 −5 0

Since A is a 2 × 2 matrix satisfying A2 + 5I = 0, then it has characteristic equation

x2 + 5 = 0, therefore trA = 0 and det A = 5. These are invariant under similarity
so any integrally similar matrix to A is of the form
Å ã
a b
c −a

where a2 + bc = −5. Amongst all integrally similar matrices to A we pick one, say
B, where |a| is minimal.
We will show that |a| = 0 or |a| = 1. Assume for contradiction that |a| > 2. Then
we must have |b| < 2a or |c| < 2a as otherwise

5 = |a2 + bc| > |bc| − a2 > 4a2 − a2 = 3a2 ,

Crux Mathematicorum, Vol. 50(7), September 2024

 OLYMPIAD CORNER /361

Å ã
1 ±1
a contradiction. If |b| < 2a, letting C = we get that CBC −1 is equal to
0 1
Å ãÅ ãÅ ã Å ãÅ ã Å ã
1 ±1 a b 1 ∓1 a±c b∓a 1 ∓1 a±c •
= = . (1)
0 1 c −a 0 1 • • 0 1 • •

Note that if a, c have opposite sign then −2 < c/a < 0, so −1 < 1 + c/a < 1 and
therefore |a + c|/|a| < 1, contradicting the minimality of |a|. If a, c have the same
sign (or c = 0) then 0 < c/a < 2, so −1 < 1 − c/a < 1 and therefore |a − c|/|a| < 1,
again a contradiction.
Å ã
0 1
Letting C = we get
1 0
Å ãÅ ãÅ ã Å ã
−1 0 1 a b 0 1 −a c
CBC = = (2)
1 0 c −a 1 0 b a
Å ã
1 0
which shows that we may assume a > 0. Letting also C = we get
0 −1
Å ãÅ ãÅ ã Å ã
1 0 a b 1 0 a −b
CBC −1 = = (3)
0 −1 c −a 0 −1 −c −a

which shows that we may assume b > 0.

If a = 0, then bc = −5 and since b > 0, then

Å ã Å ã
0 1 0 5
B= or B = .
−5 0 −1 0
Å ã
0 1
In the first case we are done while in the second case letting C = we get
−1 0
Å ãÅ ãÅ ã Å ã
−1 0 1 0 5 0 −1 0 1
CBC = =
−1 0 −1 0 1 0 −5 0

and so we are again done.

If a = 1 then bc = −6 and since b > 0, then

Å ã Å ã Å ã Å ã
1 1 1 2 1 3 1 6
B= or B = or B = or B = .
−6 −1 −3 −1 −2 −1 −1 −1

In the second case we are immediately done. In the fourth case we can use (1) to
make a = 0 contradicting the minimality of a. In the first case we can use first
(2) and then (1) to make a = 0 contradicting the minimality of a. Finally in the
third case we can use (3) to move into the second case so we are again done.

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362/ OLYMPIAD CORNER

OC670. Prove that the arithmetic sequence 5, 11, 17, 23, 29, . . . contains
infinitely many primes.
Originally from the 18th Philippine Mathematical Olympiad (2014), National Stage,
Written Phase, Problem 2.
We received 16 solutions. We present the solution by Oliver Geupel.
This follows readily from Dirichlet’s theorem on prime numbers in arithmetic pro-
gressions. We give an alternative proof in the spirit of Euclid’s famous demonstra-
tion of the infinitude of prime numbers. We are to show that there are infinitely
many primes that are congruent to -1 modulo 6. For the sake of obtaining a con-
tradiction, assume in contrary that only finitely many primes p1 , p2 , . . . , pn are
congruent to -1 modulo 6. Then,
n
Y
N= pn ≡ (−1)n (mod 6) .
k=1

Let M = N + 6 if N ≡ −1 (mod 6), and let M = N + 4 if N ≡ 1 (mod 6). Then

M is odd and coprime with N , and M ≡ −1 (mod 6).
Since every odd prime is congruent to ±1 modulo 6, the number M has a prime di-
visor that is congruent to −1 modulo 6 and that is distinct from each of p1 , . . . , pn ,
a contradiction. Consequently, there exist infinitely many primes that are congru-
ent to −1 modulo 6.

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 Yuepeng Alex Yang /363

The Sneakiest Ugly Integral Trick

Yuepeng Alex Yang
This very unintuitive integration technique was created years ago. It was intended
for integration bee competitions; however, the problems weren’t always accepted
due to the difficulty of the integrals. The trick was created out of curiosity, starting
from one definite integral, before discovering a generalized form of two different
functions. It’s definitely one of the sneakiest ways to hide an ugly quotient rule
integral.
Consider the first integral I created that inspired it all:
π
sec(x) + tan(x)
Z 3
Å ã
(sec(x) + csc(x))dx (1)
0 csc(x) + cot(x)

The standard method is Weierstrass substitution (letting u = tan( x2 ).) However,

this method can be messy: there is a much quicker way to solve this integral,
taking less than a minute.
Solution. The sneakiest trick is to make the rational function into an exponential-
log form.
π
sec(x) + tan(x)
Z 3
Å ã
(sec(x) + csc(x))dx
0 csc(x) + cot(x)
Z π
3
= eln(sec(x)+tan(x))−ln(csc(x)+cot(x)) (sec(x) + csc(x))dx
0

Let u = ln(sec(x) + tan(x)) − ln(csc(x) + cot(x)). Then du = (sec(x) + csc(x))dx.

ln(1+ √2 )
2
Z
3 ln(1+ √2 )
= eu du = [eu ]−∞ 3
= 1+ √ .
−∞ 3

A very simple solution, yet very difficult to see. So how is this done?
It combines the quotient rule and logarithmic functions. The trigonometric func-
tions inside the logarithm help hide the quotient rule in the integral, making it
look crazier than it really is. Here’s what the generalized form looks like.
ãÅ 0
f (x) f (x) g 0 (x)
Z Å ã
− dx (2)
g(x) f (x) g(x)

If you simplify it algebraically, it becomes a quotient rule integral.

ãÅ 0
f (x) g 0 (x)
Z 0
f (x) f (x)g 0 (x)
Z 0
f (x) f (x)g(x) − f (x)g 0 (x)
Z Å ã
− dx = − 2
dx = dx
g(x) f (x) g(x) g(x) g (x) g 2 (x)
(3)

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364/ The Sneakiest Ugly Integral Trick

However, with trig-identities or other functions that hide the quotient rule, it is
sometimes easier to take the u-substitution route.
ãÅ 0
f (x) g 0 (x)
Å 0
f (x) f (x) g 0 (x)
Z Å ã Z ã
− dx = eln(f (x))−ln(g(x)) − dx (4)
g(x) f (x) g(x) f (x) g(x)
Now it is easier to see that u = ln(f (x)) − ln(g(x)) solves this integral.
If the integral (1) didn’t have bounds, then the indefinite answer of the integral
would just be
sec(x) + tan(x)
+ C.
csc(x) + cot(x)
f 0 (x) g 0 (x)
The trig functions just help hide − . Let’s do some examples!
f (x) g(x)
Example 1. Consider a very famous evil integral from Calculus 2 .
1 1
Z
sec3 (x)dx = ln | sec(x) + tan(x)| + sec(x) tan(x) + C. (5)
2 2
Compute the integral:
Z π6
esec(x) tan(x) (sec(x) + tan(x)) sec3 (x)dx
0

Without knowing the trick, you might have to do a messy integration by parts.
Even WolframAlpha doesn’t know how to approach this other than numerical
approximation.
Notice that we have

sec(x) + tan(x) = eln(sec(x)+tan(x)) ,

and sec3 (x) could potentially be our du. This gives us a hint that we will be using

u = sec(x) tan(x) + ln(sec(x) + tan(x))

and du = 2 sec3 (x)dx according to equation (5). So we follow through,

Z π6
esec(x) tan(x)+ln(sec(x)+tan(x)) sec3 (x)dx
0

Let u = sec(x) tan(x) + ln(sec(x) + tan(x)). Then du = 2 sec3 (x)dx,

Z 2 √ √
1 3 +ln( 3) u 3 2 1
= e du = e3 − .
2 1 2 2
And we’re done! Quick and simple. Let’s do another example.
Example 2. Compute the integral:
Z tan−1 (x) Å
1 − 2x
ã
e
dx.
1 + x2 1 + x2

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 Yuepeng Alex Yang /365

You could possibly see that u = tan−1 (x) − ln(1 + x2 ) would solve the integral
easily. However, it is also easy for others to see that
Z tan−1 (x) Å Z tan−1 (x) −1
1 − 2x − 2xetan (x)
ã
e e
dx = dx
1 + x2 1 + x2 (1 + x2 )2
which is in the form like equation (3) where it is just a quotient rule. Hence, the
answer is Z tan−1 (x) Å −1
1 − 2x etan (x)
ã
e
dx = + C.
1 + x2 1 + x2 1 + x2

Example 3. Consider the integral:

Z
sec(x) csc(x)dx = ln | tan(x)| + C. (6)

Compute the integral:

π
sec(x) tan(x)
Z 4
(csc(x) − 1)dx.
0 sec(x) + tan(x)
This integral can be done by trig-manipulation. But observe that distributing
sec(x) with (csc(x) − 1) gives us our du.
Z π4
tan(x)
(sec(x) csc(x) − sec(x))dx
0 sec(x) + tan(x)
Z 4π

= eln(tan(x))−ln(sec(x)+tan(x)) (sec(x) csc(x) − sec(x))dx.

0

Let u = ln(tan(x)) − ln(sec(x) + tan(x)). Then du = (sec(x) csc(x) − sec(x))dx,

ln( √ 1 ) √
1
Z
2+1
eu du = √ = 2 − 1.
−∞ 1+ 2
It is definitely possible to use Weierstrass substitution or other methods to solve
this integral, but using the trick is just much faster. Here are some problems for
you to try. (I believe in you.)

Exercises
1. Compute the definite integral:
Z π2
sec3 (x)
e− sec(x) tan(x) dx.
0 sec(x) + tan(x)

2. Compute the integral:

Z π6
sin(x) + 1 cos(x) tan(x)
Å ã
− dx
0 sec(x) + 1 1 + sin(x) 1 + cos(x)

Copyright © Canadian Mathematical Society, 2024

366/ The Sneakiest Ugly Integral Trick

3. Compute the integral:

π
(sec(x) + tan(x))(x − cos(x))
Z 3
dx
π
6
x2 cos(x)

4. Compute the integral:

π
tan(x) 1 cos(x)
Z 4
Å ã
− dx
0 sin(x) + 1 sin(x) cos(x) sin(x) + 1

5. Compute the integral:

π
1 + sin(x) 1 + sin(x) cos(x) sin(x)
Z 2
Å ãÅ ãÅ ã
ln + dx
−π
2
1 + cos(x) 1 + cos(x) 1 + sin(x) 1 + cos(x)

These are the few integrals I could make so far that are neither too long nor
recognizable derivatives of quotients. See if you can find more!

Crux Mathematicorum, Vol. 50(7), September 2024

 Problems /367

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 November 15, 2024.

4961. Proposed by Michel Bataille.

Let ABC be a triangle with ∠BAC = 90◦ and BC = 3AB. Let O and G be the
circumcenter and the centroid of ∆ABC and let I be the incenter of ∆OAB.
√ Prove
that OI and BG intersect on the circumcircle of ∆AGI and that IO = 3 IA.

4962. Proposed by Leonard Giugiuc and Richdad Phuc.

Prove that if ABC is an acute angled triangle, then
cos A cos B cos C
+ + ≤ 2(cos2 A + cos2 B + cos2 C).
cos(B − C) cos(C − A) cos(A − B)

4963. Proposed by Arsalan Wares.

Two congruent semicircular arcs touch at the midpoint of diagonal BD of square
ABCD as shown. The diameters of the congruent semicircular arcs are perpen-
dicular to diagonal BD. The arcs have terminal points on the sides of square
ABCD. Points X and Y are the centers of the arcs. Point P is a terminal point,
as shown. Point Q is the point of intersection of P Y and one of the arcs, as shown.
Determine the exact value of P Q/QY .

4964. Proposed by Ovidiu Furdui and Alina Sı̂ntămărian.

Find all continuous functions f : R → R such that
Z x
f (−x) = 1 − e−t f (x − t)dt, ∀x ∈ R.
0

Copyright © Canadian Mathematical Society, 2024

368/ Problems

4965. Proposed by Ángel Plaza.

Prove that the following identities hold:
∞
(−1)n−1 1 1 1 7π 2 ln2 2
X Å ã
a) − + − ··· = − ,
n=1
n n n+2 n+4 96 8
∞
1 1 1 1 ln2 2 11π 2
X Å ã
b) − + − ··· = + ,
n=1
n n n+2 n+4 8 96

4966. Proposed by Vasile Córtoaje.

For given n ≥ 2, find the largest integer k such that
 a + a + · · · + a 2
1 2 n a2 + a22 + · · · + a2k
≥ 1
n k
for all nonnegative numbers a1 , a2 , . . . , an satisfying a1 ≤ a2 ≤ · · · ≤ an .

4967. Proposed by Marian Ursărescu.

Consider triangle ABC, where the a, b, c are lengths of the sides BC, AC, AB,
b+c
respectively. Suppose that a = , r is the inradius and R is the circumradius.
2
Prove that:
A r
sin2 ≥ .
2 2R

4968. Proposed by Dacian-Dumitru Robu.

Find all monotonic functions f : Z → Z satisfying

(x + y)f (xf (y − 1) + yf (x − 1)) = f (2xy)f (x + y − 1)

for all x, y ∈ Z.

4969. Proposed by Mihaela Berindeanu.

Let ABCD be a circumscribed quadrilateral to a circle Γ and let Ω1 , Ω2 be
the incircles of 4ABC , respectively 4ACD . Finally, let Ω1 ∩ AB = {X},
Ω1 ∩ BC = {Y } , Ω2 ∩ CD = {Z} and Ω2 ∩ AD = {T }. If XY ≡ ZT , show that
Y Z k XT .

4970. Proposed by George Apostolopoulos.

Let a, b, c be the lengths of the sides of a triangle. Prove that
ã2
1 1 1
Å Å ã
a b c
+ + ≥ 2(a + b + c) + + .
c a b a+b b+c c+a

Crux Mathematicorum, Vol. 50(7), September 2024

 Problems /369

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 novembre 2024.

4961. Soumis par Michel Bataille.

Soit ABC un triangle avec ∠BAC = 90◦ et BC = 3AB. Soit O et G le centre du
cercle et le centroı̈de de ∆ABC. Soit I le centre du circle inscrit à ∆OAB. √
Montrez
que OI et BG se coupent sur le cercle circonscrit à ∆AGI et que IO = 3 IA.

4962. Soumis par Leonard Giugiuc et Richdad Phuc.

Montrez que si ABC est un triangle à angle aigu, alors
cos A cos B cos C
+ + ≤ 2(cos2 A + cos2 B + cos2 C).
cos(B − C) cos(C − A) cos(A − B)

4963. Soumis par Arsalan Wares.

Deux arcs de cercle congruents se touchent au milieu de la diagonale BD du carré
ABCD tel qu’illustré. Les diamètres des arcs de cercle congrus sont perpendicu-
laires à la diagonale BD. Les arcs ont des points terminaux sur les côtés du carré
ABCD. Les points X et Y sont les centres des arcs. Le point P est un point
terminal, comme indiqué. Le point Q est le point d’intersection de P Y et de l’un
des arcs, comme indiqué. Déterminez la valeur exacte de P Q/QY .

4964. Soumis par Ovidiu Furdui et Alina Sı̂ntămărian.

Trouvez toutes les fonctions continues f : R → R telles que
Z x
f (−x) = 1 − e−t f (x − t)dt, ∀x ∈ R.
0

Copyright © Canadian Mathematical Society, 2024

370/ Problems

4965. Soumis par Ángel Plaza.

Montrez que les identités suivantes sont valides :
∞
(−1)n−1 1 1 1 ln2 2 7π 2
X Å ã
a) − + − ··· = − ,
n=1
n n n+2 n+4 8 96
∞
1 1 1 1 ln2 2 11π 2
X Å ã
b) − + − ··· = + ,
n=1
n n n+2 n+4 8 96

4966. Soumis par Vasile Córtoaje.

Étant donné n ≥ 2, trouvez le plus grand entier k tel que
 a + a + · · · + a 2
1 2 n a2 + a22 + · · · + a2k
≥ 1
n k
pour tous les nombres non négatifs a1 , a2 , . . . , an satisfaisant a1 ≤ a2 ≤ · · · ≤ an .

4967. Soumis par Marian Ursărescu.

Considérons le triangle ABC, où a, b et c sont les longueurs des côtés BC, AC et
b+c
AB, respectivement. Supposons que a = , et r désigne le rayon du cercle
2
inscrit tandis que R désigne le rayon du cercle circonscrit. Montrez que :
A r
sin2 ≥ .
2 2R

4968. Soumis par Dacian-Dumitru Robu.

Trouvez toutes les fonctions monotones f : Z → Z vérifiant
(x + y)f (xf (y − 1) + yf (x − 1)) = f (2xy)f (x + y − 1)
pour tout x, y ∈ Z.

4969. Soumis par Mihaela Berindeanu.

Soit ABCD un quadrilatère circonscrit à un cercle Γ et soit Ω1 , Ω2 les cercles
inscrits à 4ABC et 4ACD , respectivement. Notons Ω1 ∩ AB = {X},
Ω1 ∩ BC = {Y }, Ω2 ∩ CD = {Z} et Ω2 ∩ AD = {T }. Si XY ≡ ZT , montrez
que Y Z k XT .

4970. Soumis par George Apostolopoulos.

Soient a, b et c les longueurs des côtés d’un triangle donné. Montrez que
c 2 1 1 1
Å ã Å ã
a b
+ + ≥ 2(a + b + c) + + .
c a b a+b b+c c+a

Crux Mathematicorum, Vol. 50(7), September 2024

 Solutions /371

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 2024: 50(2), p. 82–86.

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 .
We received 14 submissions of which 13 were correct and complete. We present
the solution by Prithwijit De, slightly modified by the editor.
Denote by O the circumcentre, by H the orthocentre and by R the circumradius
of 4ABC. Extend the angle bisector of ∠BAC to meet the the circumcircle at
A0 . Denote by D the midpoint of AC.

From known properties of the nine-point circle, its centre is the midpoint of the
line segment OH and its radius is R2 . Denote by N the centre of the nine-point
circle.
Note that AO = R and AH = 2R cos ∠BAC = 2R cos 60◦ = R, so 4AHO is
isosceles. Since AH and AO are isogonal rays, ∠HAO is bisected by AA0 , so AA0
is also the perpendicular bisector of OH. Thus the midpoint N of OH lies on AA0

Copyright © Canadian Mathematical Society, 2024

372/ Solutions

and AA0 ⊥ HO. Since P T ⊥ AA0 it follows that P T ||HO. Let X be the foot of
the perpendicular from O to P T . Then T N OX is a rectangle and N T = OX.

Since ∠AT P = ∠ADP = 90◦ , AT DP is a cyclic quadrilateral. It follows that

∠T P D = ∠A0 AC = 12 ∠BAC = 30◦ , and hence in the right-angled 4P XO we
have OX = OP R R
2 = 2 . Therefore N T = 2 as well, whence T lies on the nine-point
circle. Finally, since P T ⊥ T N we conclude that P T is tangent to the nine-point
circle of 4ABC at T .

If we let P 0 be the point diametrically opposite to P on the circumcircle and T 0

be the foot of the perpendicular from P 0 to AA0 , it follows easily that N is the
midpoint of T T 0 ; hence T 0 is also on the nine-point circle and P 0 T 0 is tangent to
the nine-point circle at T 0 .

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 .

There were 12 correct solutions from 11 contributors. There was one incomplete
solution.

Editor’s Notes. In the solutions, let u, v, w denote the lengths of P A, P B, P C,

m the length of the median of triangle T to the side of length u and θ the angle
opposite to this side.

The first part, that P A, P B and P C can form a triangle, is a theorem of Dimitrie
Pompeiu. It has a very simple proof, picked up by some solvers, that uses the fact
that P A lies in the circle of centre A through B and C and the triangle inequality
for triangle P BC to obtain u < a < v + w; similarly v < w + u and w < u + v.
Another argument exploits Ptolemy’s Theorem applied to the quadrilateral ABCP
that
aw + av = AB · P C + AC · P B ≥ BC · P A = au,

for example. However, this does not tell us how to find such a triangle. A proof
that does this is provided by Brian J. McCartin on page 49 of his book Mysteries
of the equilateral triangle (Hikari, 2010). We present this proof below. A second
proof (given by three solvers) is included as a key part of Solution 1.

Determine points L, M , N on the respective sides BC, CA, AB so that P LkAB,

P M kBC and P N kCA. Then P LBN , P M AN and P M CL are equilateral trape-
zoids with base angles 60◦ whose diagonals are equal. Thus

LN = P B = v; M N = P A = u; LM = P C = w

and T has vertices L, M , N .

Crux Mathematicorum, Vol. 50(7), September 2024

 Solutions /373

A
N

P M

B L C

We present several solutions to the second part of the problem.

Solution 1, by Madhav Modak, and C.R. Pranesachar (done independently).

Consider a 60◦ clockwise rotation about C that carries A to A0 and B to B 0 = A,
as in the diagram. It takes P to P 0 . The triangle T is P AP 0 . Since P P 0 = P C
and P 0 A = P 0 B 0 = P B, the lengths of P A, AP 0 and P 0 P are respectively u, v
and w.
A = B0 A0

v
u

w θ P0
P

v
w

B C

Observe that
∠BP C = ∠AP 0 C = θ + 60◦ .
Applying the law of cosines to the triangles P BC and AP 0 P , we find that

a2 = v 2 + w2 − 2vw cos(θ + 60◦ ) and u2 = v 2 + w2 − 2vw cos θ,

whereupon
a2 + u2 = 2(v 2 + w2 ) − 2vw(cos(θ + 60◦ ) + cos θ).

Copyright © Canadian Mathematical Society, 2024

374/ Solutions

Since 4m2 = 2(v 2 + w2 ) − u2 ,

m = a/2 ⇔ a2 = 2(v 2 + w2 ) − u2
⇔ cos(θ + 60◦ ) + cos θ = 0 ⇔ (θ + 60◦ ) + θ = 180◦
⇔ θ = 60◦ .

as desired.

Solution 2, by UCLan Cyprus Problem Solving Group.

√
Without loss of generality, we may assume that a = 3 and that the triangle is
represented in the complex plane by placing the vertices A, B and C at 1, ω and
ω 2 , respectively, where ω = e2πi/3 . Let P be represented by z. Since P is in the
triangle, its real part 21 (z + z̄) exceeds −1/2.
Then

u2 = |z − 1|2 = |z|2 + 1 − z − z̄;

v 2 = |z − ω|2 = |z|2 + 1 − ω 2 z − ωz̄;
w2 = |z − ω|2 = |z|2 + 1 − ωz − ω 2 z̄.

We have

4m2 = 2(v 2 + w2 ) − u2 = 3(|z|2 + 1) + (1 − 2ω − 2ω 2 )(z + z̄)

= 3|z|2 + 3 + [3 − 2(1 + ω + ω 2 )](z + z̄) = 3 + 3(p + s),

where p = |z|2 and s = z + z̄.

√
The property m = 3/2 is equivalent to 4m2 = 3, which in turn is equivalent to
p + s = 0.
We have
v 2 + w2 − u2 |z|2 + 1 + (1 − ω − ω 2 )(z + z̄) p + 1 + 2s
cos θ = = = ,
2vw 2vw 2vw
where

v 2 w2 = (|z|2 + 1)2 − (|z|2 + 1)(z + z̄)(ω + ω 2 ) + (ω 2 z + ωz̄)(ωz + ω 2 z̄)

= (p + 1)2 + (p + 1)s + (z 2 + z̄ 2 + ω|z|2 + ω 2 |z|2 )
= (p + 1)2 + (p + 1)s + (s2 − 2p − p)
= (p + 1)2 + (p + 1)s + s2 − 3p.

The angle θ equal to 60◦ if and only if (p + s) + (s + 1) = p + 1 + 2s > 0 and

(p + 1 + 2s)2 = v 2 w2 = (p + 1)2 + (p + 1)s + s2 − 3p.

This reduces to
(s + p)(s + 1) = s(p + 1) + s2 + p = 0.

Crux Mathematicorum, Vol. 50(7), September 2024

 Solutions /375

Note that s > −1 so that s + 1 6= 0.

◦
If p + s = 0, then p + 1 + 2s
√ > 0 and θ = 60 . Conversely, θ = 60◦ implies that
◦
p + s = 0. Therefore, m = 3/2 if and only if θ = 60 .

Solution 3, by Didier Pinchon.

Taking the Cauchy-Menger determinant formula for the volume of a tetrahedron
given the side-lengths, and setting it equal to 0 (since P ABC is coplanar), we have
the equation

a4 − (u2 + v 2 + w2 )a2 + (u4 + v 4 + w4 ) − (u2 v 2 + v 2 w2 + w2 u2 ) = 0. (1)

The angle θ = 60◦ if and only if u2 = v 2 +w2 −vw. Suppose this is so. Substituting
for u2 in (1), we obtain

0 = a4 + (2v 2 − vw + 2w2 )a2 + (v 4 − v 3 w − vw3 + w4 )

= (a + v − w)(a − v + w)(a2 − v 2 − vw − w2 ).

From the triangle inequality applied to triangle P BC, the first two factors are
nonzero, so a2 = v 2 + vw + w2 .
On the other hand, suppose that a2 = v 2 + vw + w2 . Then the equation becomes

0 = (u2 − v 2 + vw − w2 )[u2 − (v + w)2 ].

Since u < v + w, then u2 = v 2 + w2 − vw and θ = 60◦ .

The median m = a/2 if and only if u2 = 2v 2 + 2w2 − a2 . Substituting this into
(1) yields

0 = 3[a4 −2(v 2 +w2 )a2 +(v 4 +v 2 w2 +w4 )] = 3(a2 −v 2 +vw−w2 )(a2 −v 2 −vw−w2 ).

Because P is inside the triangle ABC, ∠BP C exceeds 60◦ and so its cosine is less
than 1/2. Hence

a2 − v 2 + vw − w2 > a2 − (v 2 + w2 − 2vw cos θ) = 0.

Hence a2 = v 2 + vw + w2 and so θ = 60◦ .

Conversely, if θ = 60◦ , then both a2 = v 2 + w2 + vw and u2 = v 2 + w2 − vw.
Adding these gives a2 + u2 = 2(v 2 + w2 ) and so m = a/2.

Solution 4, by Theo Koupelis.

Let β = ∠AP B and γ = ∠AP C. Then ∠BP C = 360◦ − (β + γ) and

u2 + v 2 − a2 u2 + w2 − a2 v 2 + w2 − a2
cos β = , cos γ = , cos(β + γ) = .
2uv 2uw 2vw

Copyright © Canadian Mathematical Society, 2024

376/ Solutions

Squaring the equation sin β sin γ = cos β cos γ − cos(β + γ) leads to

0 = a4 − (u2 + v 2 + w2 )a2 + (u4 + v 4 + w4 ) − (u2 v 2 + v 2 w2 + w2 u2 )

= (a2 − v 2 − w2 − vw)2 + (v + w + u)(v + w − u)(a2 − u2 − 2vw). (1)

Since 4m2 = 2(v 2 + w2 ) − u2 , m = a/2 if and only if a2 = 2v 2 + 2w2 − u2 . The

angle θ is equal to 60◦ if and only if u2 = v 2 + w2 − vw. Consider the following
two identities:

[a2 − 2(v 2 + w2 ) + u2 ] − [u2 − v 2 − w2 + vw] = a2 − v 2 − w2 − vw;

[a2 − 2(v 2 + w2 ) + u2 ] − 2[u2 − v 2 − w2 + vw] = a2 − u2 − 2vw.

Suppose that m = a/2. Then a2 − 2(v 2 + w2 ) + u2 = 0 and, using the identities,

we can render equation (1) as

0 = [u2 − v 2 − w2 − vw][(a2 − v 2 − w2 − vw) + 2((v + w)2 − u2 )]

= [u2 − v 2 − w2 − vw][(a2 − u2 ) + ((v + w)2 − u2 ) + vw].

The second factor being positive, we must have u2 = v 2 + w2 − vw and θ = 60◦ .

Suppose that θ = 60◦ . Then u2 − v 2 − w2 + vw = 0 and we can render (1) as

0 = [a2 − 2(v 2 + w2 ) − u2 ][(a2 − v 2 − w2 − vw) + (v + w)2 − u2 ]

= [a2 − 2(v 2 + w2 ) − u2 ][(a2 − u2 ) + vw].

Again, the second factor is positive, and so a2 = 2(v 2 + w2 ) − u2 and m = a/2.

Editor’s Comments. The equation linking a, u, v, w can be written in a comely

symmetrical form

(u2 + v 2 + w2 + a2 )2 = 3(u4 + v 4 + w4 + a4 ).

Vivek Mehra drew attention to Crux Problem 3832, whose solution appears in
Crux 40:4 (April, 2014), 168-171, in which it is shown that, if P is within an
equilateral triangle of side 1, then

1 + u4 + v 4 + w4 = u2 + v 2 + w2 + u2 v 2 + v 2 w2 + w2 u2 .

This formula also applies when P is outside the triangle (this can be argued from
the Cauchy-Menger determinant formula). Thus, taking a = 1, and letting D be
the reflection of A in BC, the quadratic

t2 − (1 + v 2 + w2 )t + (v 4 + w4 − v 2 − w2 − v 2 w2 )

has two roots, u2 and z 2 = P D2 . Thus u2 + z 2 = 1 + v 2 + w2 .

Crux Mathematicorum, Vol. 50(7), September 2024

 Solutions /377

Suppose that θ = 60◦ . Then, as in Solution 1, ∠BP C = 120◦ . Since ∠BDC = 60◦ ,
the quadrilateral BDCP is concyclic. By Ptolemy’s theorem, z = v + w; hence
z 2 = v 2 + w2 = 2vw.
Thus, P belongs to the circumcircle of triangle BCD, where D is the reflection
of A in BC. This is the dilatation with factor 2 about Q of the incircle of this
triangle, where Q is the reflection of O in BC.
Since θ = 60◦ , then u2 = v 2 + w2 − vw. Now

1 + v 2 + w2 − u2 = z 2 = v 2 + w2 + 2vw,

whence 2vw = 1 − u2 . Therefore u2 = v 2 + w2 − 21 (1 − u2 ) from which

1 = 2(v 2 + w2 ) − u2 = 4m2 .

For the converse, Mehra showed that if m = 1/2, then BDCP must be concyclic.
This was done by assuming z > v + w and obtaining a contradiction.
We close with the conjecture that under the hypotheses of the problem except that
P is outside of ABC, then the triangle has a median equal to a/2 if and only if
the corresponding angle is 120◦ .

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).
We received 28 submissions, 26 of which were correct and complete. We present
the solution submitted by the Eagle Problem Solvers, which is representative of the
majority of solutions.
tx2
We assume f (7) 6= 0, since otherwise the fraction f (7) would be undefined. Making
tx2
the substitution u = f (7) , we see that

xf (7) x3
tx2 f (7)
Z Å ã Z
f dt = 2 · f (u) du = 3f (7)x4 ,
0 f (7) x 0

and hence Z x3
f (u) du = 3x6 .
0
This is satisfied when x = 0, so we assume x 6= 0. Since f is continuous, we use
Part I of the Fundamental Theorem of Calculus to differentiate both√sides with
respect to x, getting f (x3 ) · 3x2 = 18x5 and f (x3 ) = 6x3 . With x = 3 7, we see
that f (7) = 6 · 7 = 42.

Copyright © Canadian Mathematical Society, 2024

378/ Solutions

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 .

There were 19 correct solutions from 17 solvers, and three incomplete solutions.
We present 3 solutions.
Solution 1, by José Luis Arregui and UCLan Cyprus Problem Solving Group (done
independently).
Setting x = 0 yields
f (y + 1) = f (y) + 1
for y ≥ 0. When y = 0, we obtain

f (x2 + 1) = xf (x) + f (0) + 1 = xf (x) + f (1)

for x ≥ 0. In particular, f (2) = 2f (1). From (x, y) = (0, 1), f (2) = f (1) + 1, so
that f (1) = 1.
Since
f (x2 ) + 1 = f (x2 + 1) = xf (x) + f (1) = xf (x) + 1,
f (x2 ) = xf (x) and
√ √
f (x + y) = f (x + y + 1) − 1 = xf ( x) + f (y) = f (x) + f (y)

for x, y ≥ 0. Therefore

f ((x + 1)2 ) = f (x2 + x + x + 1) = f (x2 ) + 2f (x) + f (1) = f (x2 ) + 2f (x) + 1.

But

f ((x + 1)2 ) = (x + 1)f (x + 1) = (x + 1)(f (x) + 1) = xf (x) + x + f (x) + 1.

Comparing the two equations shows that f (x) = x for x ≥ 0.

Note that we did not require the hypothesis that f (x) be increasing.

Solution 2, by Raymond Mortini and Rudolf Rupp.

As in Solution 1, we can show that f (1) = 1, f (x + 1) = f (x) + 1 and that
f (x) = x1/2 f (x1/2 ) for x ≥ 0. Therefore, when x > 1 and n is a positive integer,
n
f (x) f (x1/2 ) f (x1/2 )
= = · · · = .
x x1/2 x1/2n
Since f (x) is increasing, we can find the right-hand limit at 1:

lim f (x) = c ≥ f (1),

x↓1

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 Solutions /379

where c = inf{f (x) : x ≥ 1}. Therefore

n
f (x) f (x1/2 )
= lim n = c.
x n→∞ x1/2

Thus f (x) = cx for x > 1.

Since 1 = f (1) = f (2) − 1 = 2c − 1, it follows that c = 1. When x > 0,
f (x) + 1 = f (x + 1) = x + 1,
so that
f (x) = f (x + 1) − 1 = (x + 1) − 1 = x.
This is also true for x = 0. Therefore f (x) = x for all x ≥ 0.

Solution 3, by Mehra Vivek.

As in Solution 1, we can prove that f (x + y) = f (x) + f (y), f (x2 ) = xf (x) and
f (1) = 1. It is known that therefore f (rx) = rx for all x ≥ 0 and positive rationals
r. Let 0 < t < 1 and (x, y) = (t, 1 − t2 ) where 0 < t < 1. Then
2 = f (2) = tf (t) + f (1 − t2 ) + 1 = f (t2 ) + f (1 − t2 ) + 1,
whence f (t2 ) + f (1 − t2 ) = 1. Let x > 0 and m, n be positive integers.
Suppose that t2 = m/(n + m). Then
n n
1 − t2 = = t2 .
m+n m
Suppose, if possible, that f (t2 ) > t2 . Then
m m m
f (t2 ) = f (1 − t2 ) = (1 − f (t2 )) < (1 − t2 ) = t2 ,
n n n
a contradiction. Therefore f (r) = r for all positive rationals r. Since f (x) is
monotonic, f (x) is at once equal to the supremum of its values on [0, x) and
infinimum of its values on (x, ∞), namely x.

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→∞

We received 15 submissions, 10 of which were correct and complete. We present

the solution submitted by the proposer, which is representative of the majority of
solutions.
First, we have
∞ Z 1
X (−1)k+1 xn+k
Sn = dx
0 k
k=1

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380/ Solutions

with
∞ Z 1 ∞
(−1)k+1 xn+k 1 1 1
X X Å ã
dx = <∞ since ≤ 2 ,
0 k k(k + n + 1) k(k + n + 1) k
k=1 k=1

hence we can change the order sum/integral and obtain

∞
Z 1 X ! Z 1
(−1)k+1 xn+k
Sn = dx = xn ln(1 + x) dx.
0 k 0
k=1

Integrating by parts yields

1 ò1
xn 1 1 n
Z ï Z Å ã
x
xn ln(1 + x) dx = · x ln(1 + x) − x ln(1 + x) + dx,
0 n 0 n 0 1+x

that is,
ln(2) 1 1
Sn = − · Sn − · In (1)
n n n
Z 1
x
where In = xn ·
dx.
0 1+x
We know that if f is continuous on [0, 1], then
Z 1
lim n xn f (x) dx = f (1).
n→∞ 0

It follows that
ln(2)
Sn = + o(1/n)
n
and
1
In = + o(1/n),
2n
so that (1) gives
ln(2) 2 ln(2) + 1
− Sn = + o(1/n2 )
n 2n2
as n → ∞. By integration by parts again, we have
ò1 1
xn+2 1 x2 + 2x
ï Z
In = − xn · dx.
n(1 + x) 0 n 0 (x + 1)2

Hence
1 3
In = − + o(1/n2 )
2n 4n2
and from (1) again, we obtain

ln(2) 1 ln(2) 2 ln(2) + 1 1 1 3

Å ã Å ã
2 2
Sn = − − + o(1/n ) − − + o(1/n ) .
n n n 2n2 n 2n 4n2

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 Solutions /381

Finally, we readily obtain n3 Sn = an2 + bn + c + o(1) with

1 + 2 ln(2) 5 + 4 ln(2)
a = ln(2), b = − , c= .
2 4

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.

We received 25 correct and complete submissions, out of which we present the one
by Walther Janous, lightly edited.
A

10
B E H F C

Let s be the length of AB and D, E, F , G, H vertices of the trapezoids as in

the diagram. Since every section in the diagram has equal area, we obtain the
following relations:
2
[DEF ] = [ABC]
3
and
1
[F GH] = [ABC].
2

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382/ Solutions

√
s s 2
Since DEF and F GH are equilateral triangles, we have HF = √ and EF = √ .
2 3
This in turn leads to
Å√ √ ã
2 1 2− 3
ã Å
10 = EF − HF = s √ − √ =s √
3 2 6

and thus √
10 6 √ √ √ √
s= √ = 10 6(2 + 3) = 20 6 + 30 2.
2− 3

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.

We received 19 submissions, all of which were correct. Most of the solutions were
based on one of two strategies, so we will exhibit an example of each approach.
Solution 1 is a composite of the solutions that follow from the fact that the diagonals
of a parallelogram intersect at their common midpoint.

Let I be the incenter of ABC. We first argue that the quadrilateral F IES is a
rhombus: Triangle AF E is isosceles (with AF = AE as tangents to the incircle)
and therefore the altitude AS is the perpendicular bisector of F E and the angle
bisector of ∠F AE. But I is equidistant from E and F , so it too lies on AS,

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 Solutions /383

whence F IES is symmetric about IS. But it is also symmetric about F E because
∠IF E = ∠EF S — they are both equal to half ∠F AE because corresponding
sides are perpendicular, namely,

IF, F E ⊥ F A, AI, and EF, F S ⊥ IA, AE.

It follows that opposite sides of F IES are equal and parallel, as claimed. Likewise,
opposite sides of T DIF are equal and parallel. It follows that the segments ES
and DT are equal and parallel. Consequently, EST D is a parallelogram, and
we deduce that its diagonals T E and SD have the same midpoint. In the same
way, U F and SD have the same midpoint, and we conclude that SD, T E and U F
concur at their common midpoint.
The use of vectors here yields further information: That common midpoint, call
it P , is the nine-point center of triangle DEF . To see this, note that from the
−
→ −→ −→
rhombus F IES we have IS = IE + IF . It follows that the vector from I to the
midpoint of DS satisfies
−→ 1 Ä−→ − →ä 1 Ä−→ −→ −→ä
IP = ID + IS = ID + IE + IF . (1)
2 2
−→ −→ −→
We recognize ID + IE + IF to be the vector from the triangle’s circumcenter to its
orthocenter. Since the triangle’s nine-point center is halfway between those other
two centers, equation (1) tells us that P is the nine-point center, as claimed.
Editor’s Comments. While many submissions obtained a version of equation (1),
only Corneliu Manescu-Avram and Marie-Nicole Gras noted the role of the nine-
point center.

Solution 2 is a composite of the solutions based on Ceva’s theorem.

A
We saw in solution 1 that ∠IF E = 2; similarly, ∠T F I = B; consequently,
A
∠T F E = ∠T F I + ∠IF E = + B.
2

By the Sine Rule applied to triangle T F E we have sin (∠F

FT
ET )
= sin (∠T F E)
TE , so
that Å ã
A FT
sin (∠F ET ) = sin +B · .
2 TE
Using the analogous expressions for sin (∠T ED), sin (∠EDS), sin (∠SDF ), sin (∠DF U ),
sin (∠U F E) and the fact that T D = T F, SE = SF, U E = U D we get

sin (∠F ET ) sin (∠EDS) sin (∠DF U ) sin ( A2 + B) sin ( C2 + A) sin ( B2 + C)

· · = · · .
sin (∠T ED) sin (∠SDF ) sin (∠U F E) sin ( C2 + B) sin ( B2 + A) sin ( A2 + C)
We now observe that
Å ã Å ã
A A
+B + + C = 180◦ ,
2 2

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384/ Solutions

thus Å ã Å ã
A A
sin + B = sin +C .
2 2
Likewise, we have
Å ã Å ã Å ã Å ã
B B C C
sin + C = sin +A and sin + A = sin +B .
2 2 2 2

Therefore
sin (∠F ET ) sin (∠EDS) sin (∠DF U )
· · = 1.
sin (∠T ED) sin (∠SDF ) sin (∠U F E)
Because the cevians along SD, T E and U F are all inside ∆DEF , they cannot be
parallel; consequently, by the trigonometric version of Ceva’s theorem we conclude
that the lines SD, T E and U F concur.

4918. Proposed by Yagub Aliyev.

2
λx
Let L = limλ→+∞ R b 2 .
a
λt dt
a) Show that if 0 ≤ a ≤ x < b, then L = 0.
b) Show that if 0 ≤ a < x = b, then L = +∞.
We received 16 correct and 3 incomplete solutions. We present the solution by
Raymond Mortini and Rudolf Rupp.
Let 0 ≤ a < b.
a) Suppose that a ≤ x < b. Then for a ≤ t ≤ b we have 1 ≥ t/b, and so, for λ > 1,
2 2 2 2 2
lx blx 2b(log l) lx 2b(log l) lx −b
0 ≤ I(l) := R b 2 ≤ = 2 2 = .
1 − la2 −b2
Rb
lt dt tlt2 dt l −l
b a
a a

Since
2
−b2 ) log l
(log l)e(log l)(x = → 0 as l → ∞,
e(log l)(b2 −x2 )
we have that liml→∞ I(l) = 0.

b) Suppose that x = b. Then for a ≤ t ≤ b we have 1 ≤ t/a and so

2 2 2 2
lb alb 2a(log l) lb 2a(log l) lb
I(l) = R b ≥ R b 2 = b2 2 ≥ = 2a log l → ∞ as l → ∞.
2
lt dt tl t dt l −l a l b2
a a

Editor’s Comments. Theo Koupelis and G. C. Greubel independently proposed

a short solution using asymptotic behaviour of imaginary error function erfi(x).
Walther Janos, Oliver Geupel, and several other solvers contributed proofs in the
spirit of classical analysis with arbitrarily small ’s and split integrals.

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 Solutions /385

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.

We received 6 submissions and they were all correct. We present 2 different solu-
tions by the majority of solvers.
Solution 1.
Let ω = exp(2πi/3); then we have ω 3 = 1, 1 + ω + ω 2 = 0, ω = ω 2 . Observe that
I6 + ω 2 A + ωB = I6 + ωA + ω 2 B; it follows that

det(I6 + ωA + ω 2 B) det(I6 + ω 2 A + ωB)

= det(I6 + ωA + ω 2 B) · det(I6 + ωA + ω 2 B)
= | det(I6 + ωA + ω 2 B)|2 ≥ 0.

On the other hand, by the hypothesis, we calculate

(I6 + ωA + ω 2 B)(I6 + ω 2 A + ωB)

= I6 + A2 + B 2 + (ω + ω 2 )(A + B) + ωBA + ω 2 AB
= AB + A + B − (A + B) + ωBA + ω 2 AB = ω(BA − AB).

Now the conclusion follows from taking the determinant on both sides of the above
equation.

Solution 2. Write I = I6 . Let X = A − I and Y = B − I. Then the hypothesis

implies that

(X + I)2 + (Y + I)2 = (X + I)(Y + I) + (X + I) + (Y + I) − I,

that is, X 2 + Y 2 = XY . Furthermore, note that BA − AB = Y X − XY , and

(X + Y )2 + 3(X − Y )2 = 4(X 2 + Y 2 ) − 2(XY + Y X) = 2(XY − Y X) .

Thus,

det((X + Y )2 + 3(X − Y )2 )
det(BA − AB) =
(−2)6
√ √
det(X + Y + i 3(X − Y )) det(X + Y − i 3(X − Y ))
=
26
√ 2
det(X + Y − i 3(X − Y )
= ≥0
26

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386/ Solutions

4920. Proposed by Ángel Plaza.

Z 1
log(1 + xk + x2k + · · · + xnk )
If k > 1 and n ∈ N, evaluate dx.
0 x
We received 19 submissions, 18 of which were correct. We present the joint solution
by Ulrich Abel and Vitaliy Kushnirevych.
Using
1
log(1 − xp ) π2
Z
dx = − , p > 0, (1)
0 x 6p
one obtains
Z 1 Z 1
log(1 + xk + x2k + · · · + xnk ) log(1 − xk(n+1) ) − log(1 − xk )
dx = dx
0 x 0 x
π2 1 π2 n
Å ã
=− −1 = .
6k n + 1 6k(n + 1)
P∞ tm
To prove the equation (1), we use log(1 − t) = − m=1 m , so we have for p > 0
1 ∞ Z 1 pm−1 ∞ 1 ∞
log(1 − xp ) 1 xpm 1 π2
Z X x X X
dx = − dx = − =− = − .
0 x m=1 0
m m=1
m pm 0 m=1
pm2 6p

Crux Mathematicorum, Vol. 50(7), September 2024