Volume/tome 47, issue/numéro 8

October/octobre 2021
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:
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cycle universitaire publiée par la Société mathématique du Canada. Principalement de nature éducative,
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Supported by / Soutenu par :

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Editorial Board

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

MathemAttic Editors John McLoughlin University of New Brunswick

Shawn Godin Cairine Wilson Secondary School
Kelly Paton Quest University Canada
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 Berlin, Germany
Magdalena Georgescu BGU, Be’er Sheva, Israel
Chip Curtis Missouri Southern State University
Philip McCartney Northern Kentucky University

Guest Editors Yagub Aliyev ADA University, Baku, Azerbaijan

Ana Duff Ontario Tech University
Andrew McEachern York University
Vasile Radu Birchmount Park Collegiate Institute
Aaron Slobodin University of Victoria
Chi Hoi Yip University of British Columbia
Samer Seraj Existsforall Academy
Translators Rolland Gaudet Université de Saint-Boniface
Jean-Marc Terrier University of Montreal
Editor-at-Large Bill Sands University of Calgary
Managing Editor Denise Charron Canadian Mathematical Society
 IN THIS ISSUE / DANS CE NUMÉRO

371 Editorial Kseniya Garaschuk

372 MathemAttic: No. 28
372 Problems: MA136–MA140
374 Solutions: MA111–MA115
377 Olympiad Corner: No. 396
377 Problems: OC546–OC550
379 Solutions: OC521–OC525
384 An introduction of the problem of finding the chromatic num-
ber of the plane (II) Veselin Jungić
392 Problems: 4671–4680
397 Solutions: 4621–4630

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 /371

EDITORIAL
I enjoy a variety of hand crafts and, being a mathematician, I ponder about math
in each of them. I scrapbook and realize how much proportional thinking goes into
photography and design work. I knit and connections to modular arithmetic are
obvious each time I have to follow an intricate pattern. I sew and wonder how it is
possible that my grandmother was such an amazing natural topologist. Recently,
I also picked up embroidery. I am originally from Belarus, where national clothing
and linens are known to be adorned with beautiful red and white embroidered
designs. Various mathematical elements in standard patterns are clear: rotational
and reflection symmetries, parity details, features occurring in multiples and so on.
Different elements mean different things, such as beauty, land, sun, soul, spring.
But having finished my first piece of non-traditional embroidery (I figured I’ll
practice on something more playful before moving onto patterns of cultural signif-
icance), I discovered a deeper connection to the approach and the presentation of
mathematics. Indulge me while I philosophize.
The front of the finished piece is a neatrly “drawn” pattern; like a finished math-
ematical solution or proof, it looks well thoughtout and nicely laid out. But the
reverse side of the embroidery tells a different story: stitches overlap in a disor-
ganized fashion, there are knots for each thread’s beginning and end, there are
straight lines connecting seemingly unrelated points on the pattern. The overall
pattern is still recognizable, but I wouldn’t boast about my skills if this was the
final product. This side is the messy story of creation that we see so seldom in
math. This is unsuccessful solution attempts, trial and error approaches, small ex-
amples, ideas that never worked out, proof write-ups that were deemed not elegant
enough; this is draft work that noone but the craftsperson themself sees.

Every presented solution in Crux makes the work look effortless and immediate,
like the person knew what they were doing from the very first to the last step
with no additional investigations, miscalculations, wrong turns. But do not be
fooled: there is the reverse side to each solution, where the author might have
laboured for hours before arriving at the idea and producing a clean write-up.
Noone’s first draft is their final draft. Practice will make this process faster:
expert problem solvers recognize familiar patterns, know shortcuts and are more
proficient in writing. But no matter how expert, the reverse side is never as perfect
as the front. Though it is necessary: the imperfections make the pattern possible.
Kseniya Garaschuk

Copyright © Canadian Mathematical Society, 2021

372/ MathemAttic

MATHEMATTIC
No. 28
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 December 30, 2021.

MA136. Proposed by Ed Barbeau.

Determine all sets consisting of an odd number 2m + 1 of consecutive positive
integers, for some integer m ≥ 1 such that the sum of the squares of the smallest
m + 1 integers is equal to the sum of the squares of the largest m integers.

MA137. Triangle ABC has area 1. X, Y are points on the side AB and Z
a point on the side AC such that XY = 2AX, XZ is parallel to Y C and Y Z is
parallel to BC. Determine the area of triangle XY Z.

MA138. Proposed by Aravind Mahadevan.

Prove that csc 6◦ + csc 78◦ − csc 42◦ − csc 66◦ = 8.

MA139. The shape below was created by pasting together 25 unit squares.
When a similar shape is created with n squares, its perimeter is 100 units. Deter-
mine n.

MA140.
If f (x) = 1 − x − x3 , what are all real values of x which satisfy
3
1 − f (x) − (f (x)) > f (1 − 5x)?

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 MathemAttic /373

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 30 decembre 2021.

MA136. Proposé par Ed Barbeau.

Déterminer tous les ensembles comprenant un nombre impair 2m + 1 d’entiers
positifs consécutifs, où m est un entier, m ≥ 1, sachant qu’en plus la somme des
carrés des m + 1 plus petits éléments de l’ensemble égale la somme des carrés des
m plus grands éléments.

MA137. Le triangle ABC a une surface de 1. Les points X et Y se trouvent

sur le côté AB, tandis que le point Z se trouve sur le côté C, de façon à ce
que XY = 2AX, que XZ soit parallèle à Y C et que Y Z soit parallèle à BC.
Déterminer la surface du triangle XY Z.

MA138. Proposé par Aravind Mahadevan.

Démontrer que csc 6◦ + csc 78◦ − csc 42◦ − csc 66◦ = 8.

MA139. La forme présentée ci-bas a été créée à l’aide de 25 carrés de côté

unitaire. Lorsqu’une forme similaire est créée à l’aide de n carrés, son périmètre
est de 100 unités. Déterminer n.

MA140. Si f (x) = 1 − x − x3 , déterminer toutes les valeurs réelles x telles

que
3
1 − f (x) − (f (x)) > f (1 − 5x)?

Copyright © Canadian Mathematical Society, 2021

374/ MathemAttic

MATHEMATTIC
SOLUTIONS
Statements of the problems in this section originally appear in 2021: 47(3), p. 121–122.

MA111. Peter and Steve had a 50-metre race. When Peter crosses the finish
line, Steve was 5 metres behind. A rematch is set up, with Peter handicapped by
being 5 metres behind the starting line. Assuming that the boys run at the same
speed as before, who wins the race (or is it a tie) and by how much?
Originally from 2000 ESSO-CMS Math Camp, Problem Set 4, problem 5.
We received 6 submissions, of which 5 were correct and complete. We present the
solution by Dominique Mouchet, Lycée Evariste de Parny, St Paul, France.
Notons vp et vs les vitesses de Peter et Steve. Soit t la durée de la première course.
Alors vp = 50/t et vs = 45/t. D’où vp /vs = 10/9. Lors de la 2e course, Steve a
couru pendant une durée ts = 50/vs et Peter une durée

55 55 49, 5
tp = = 10 = < ts .
vp 9· vs vs

Donc Peter arrive encore le premier.

Pendant le temps tp , Steve a parcouru vs × tp = 49, 5 m. Peter a donc 50 cm
d’avance.

MA112. Circle C1 has radius 13 and is tangent to line ` at Y . Circle C2 has

radius 23, is tangent to ` at Z and is externally tangent to C1 at X. The line
through X and Y intersects C2 at P . Determine the length of P Z.
Inspired by 1983 Euclid contest, essay question 2b.
We received 5 submissions, all of which were correct. We present the solution by
Miguel Amengual Covas, modified by the editor.
P

O2

O1 X

Y Z

Crux Mathematicorum, Vol. 47(8), October 2021

 MathemAttic /375

Let O1 and O2 be centres of C1 and C2 , respectively. The line O1 O2 , joining

the centres of the touching circles, goes through the point X. Observe that
∆O1 Y X and ∆O2 P X are isosceles triangles. As ∠O2 XP = ∠O1 XY , we have
that ∠XO1 Y = ∠XO2 P and P O2 is parallel to O1 Y .
As O1 Y is perpendicular to l, so is P O2 . As O2 Z is also perpendicular to l, it
follows that the length of P Z is the diameter of C2 . Therefore, P Z = 46.

MA113. For any positive number t, btc denotes the integer part of t and
{t} denotes the “decimal” part of t. If x + {y} = 7.32 and y + bxc = 8.74, then
determine {x}.
Originally from 2000 ESSO-CMS Math Camp, Problem Set 3, problem 3.
We received 8 correct solutions for this problem. We present the solution by the
Missouri State University Problem Solving Group.
There are integer n and m such that n ≤ x < n + 1 and m ≤ y < m + 1. Then
bxc = n and {y} = y − m. The two given equations can be written as

x + y − m = 7.32,
y + n = 8.74.

Subtracting, we get

x − n − m = −1.42, or x + 1.42 = n + m.

Since n + m is an integer, we must have {x} = 0.58.

Remark. Since 0 ≤ {y} < 1, we have 6.32 ≤ x < 7.32, hence x = 6.58 and
y = 2.74.

MA114. Let p, q, and r be positive constants. Prove that at least one of the
following equations has real roots.

px2 + 2qx + r = 0
rx2 + 2px + q = 0
qx2 + 2rx + p = 0

Originally from 1983 Descartes Contest, problem 6b.

We received 9 correct solutions and one incorrect solution. Both solutions below
assume only that p, q, r are real.
Solution 1, by Vikas Kumar Meena and Dominique Mouchet (done independently).
Adding the discriminants of the three equations, we get

4(q 2 − pr) + 4(p2 − rq) + 4(r2 − qp) = 2[(p − q)2 + (q − r)2 + (r − p)2 ] ≥ 0,

Copyright © Canadian Mathematical Society, 2021

376/ MathemAttic

whence at least one of the discriminants is nonnegative. The corresponding equa-

tion has real roots.

Solution 2, by Corneliu Manescu-Avram, Daniel Vacaru, and the Missouri State

University Problem Solving Group (done independently).
If the discriminants of all three equations are negative, then

q 2 < pr, p2 < rq, r2 < qp,

whence
p2 q 2 r2 < (pr)(rq)(qp) = p2 q 2 r2 ,
which is a contradiction. Hence at least one equation must have real roots.

Comments from the editor. One solver assumed, wolog, that p ≤ q ≤ r. This
is fine, but requires a bit of argument. His conclusion that r2 − pq ≥ 0 works
when p, q, r > 0, but not in general. A better approach would be to assume,
wolog, that r, say, has largest absolute value, whereupon r2 − pq ≥ 0 and we can
identify which equation has real roots. Conversely, if p, q, r all have the same sign,
a similar argument (or one along the lines of Solution 2), will ensure that at least
one equation has nonreal roots. When p, q, r do not all have the same sign, then
two equations must have real roots and either possibility obtains for the third.

MA115. I met a person the other day that told me they will turn x years old
in the year x2 . What year were they born?
Inspired by 1982 Fermat Contest, problem 19.
We received 8 submissions, all correct. We present the solution by Bellamy Kas-
tanya of Garth Webb Secondary School in Oakville, Ontario, Canada.
Since the birthday will happen in the future, we are interested in the year in the
near future which is a square. There are two candidates: 2025 and 2116.
If it is 2025, then the person will turn 45. This means that the person was born in
1980. If it is 2116, then the person will turn 46. This means that the person will
be born in 2070 (so, I cannot meet this person yet). So, the only logical solution
is the first one and the person was born in 1980.

Crux Mathematicorum, Vol. 47(8), October 2021

 OLYMPIAD CORNER /377

OLYMPIAD CORNER
No. 396
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 December 30, 2021.

OC546. Let a, b, c be three real numbers. For each positive integer n,

an + bn + cn is an integer. Prove that there exist three integers p, q, r such that
a, b, c are the roots of the equation x3 + px2 + qx + r = 0.

OC547. In a non-equilateral triangle ABC, I is the incentre and O is the

circumcentre. Prove that ∠AIO ≤ 90◦ if and only if 2BC ≤ AB + AC.

OC548. Each of the digits 1, 3, 7, 9 occurs at least once in the decimal

representation of some positive integer. Prove that one can permute the digits of
this integer such that the resulting integer is divisible by 7.

OC549. In the triangle ABC we have AB = AC and ∠BAC = 90◦ . Consider

points M and P on AB such that AM = BP . Let D be the midpoint of BC, R a
point on CM , and Q a point on BC such that A, R, Q are collinear and the line
AQ is perpendicular to CM . Show that:
(a) ∠AQC ∼
= ∠P QB;
(b) ∠DRQ = 45◦ .

OC550. The real numbers x, y, and z are not all equal and satisfy:
1 1 1
x+ = y + = z + = k.
y z x
Determine all possible values of k.

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Copyright © Canadian Mathematical Society, 2021

378/ 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 30 decembre 2021.

OC546. Soient a, b et c trois nombres réels tels que an + bn + cn est un entier

pour tout n entier positif. Démontrer qu’il existe trois entiers p, q et r de façon à
ce que a, b et c soient les trois racines de x3 + px2 + qx + r = 0.

OC547. Dans un triangle ABC qui n’est pas équilatéral, soit I le centre du
cercle inscrit et O le centre du cercle circonscrit. Démontrer que ∠AIO ≤ 90◦ si
et seulement si 2BC ≤ AB + AC.

OC548. Chacun des chiffres 1, 3, 7 et 9 a lieu au moins une fois dans la

représentation décimale d’un certain entier positif. Démontrer qu’il et possible de
permuter les chiffres de cet entier de façon à ce que l’entier résultant soit divisible
par 7.

OC549. Un triangle ABC est tel que AB = AC et ∠BAC = 90◦ . Soient M

et P des points dans AB tels que AM = BP ; soit aussi D le point milieu de BC;
enfin, soit R un point dans CM et Q un point dans BC tels que A, R et Q sont
allignés et que la ligne AQ est perpendiculaire à la ligne CM . Démontrer que:
(a) ∠AQC ∼
= ∠P QB;
(b) ∠DRQ = 45◦ .

OC550. Les nombres réels x, y et z ne sont pas tous égaux. Ils vérifient
1 1 1
x+ = y + = z + = k.
y z x
Déterminer toute valeur possible de k.

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 OLYMPIAD CORNER /379

OLYMPIAD CORNER
SOLUTIONS
Statements of the problems in this section originally appear in 2021: 47(3), p. 135–136.

OC521. In the plane there are two identical circles with radius 1, which are
tangent externally. Consider a rectangle containing both circles, each side of which
touches at least one of them. Determine the largest and the smallest possible area
of such a rectangle.
Originally from 2018 Czech-Slovakia Math Olympiad, 3rd Problem, Category A,
First Round.
We received 8 solutions. We present 2 solutions.
Solution 1, by UCLan Cyprus Problem Solving Group.

From the figure above we see that the area is equal to A = (2 + x)(2 + y) where
x, y satisfy x2 + y 2 = (O1 O2 )2 = 4.
By Cauchy-Schwarz, we have
» x2 + y 2 √
A = 4 + 2(x + y) + xy 6 4 + 2 2(x2 + y 2 ) + =6+4 2
2
√
with equality if and only if x = y = 2.
p
Since also x + y > x2 + y 2 = 2, then
1
A = 4 + 2(x + y) + xy = 2 + (x + y)2 + 2(x + y) > 8
2
with equality if and only if x = 2, y = 0 or x = 0, y = 2.
Both equality cases can actually be achieved but we omit the figures.

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380/ OLYMPIAD CORNER

Solution 2, by the Missouri State University Problem Solving Group.

We will assume that the bounding rectangle is oriented so that its sides are parallel
to the coordinate axes and its center is at the origin. Suppose the two circles are
tangent at the origin, arranged as follows. One circle has center (cos t, sin t),
and the other circle has center (− cos t, − sin t), for 0 ≤ t ≤ π/2. The bounding
rectangle has dimensions 2(1 + cos t) and 2(1 + sin t). We wish to find the largest
and smallest values of the area function A(t) = 4(1+cos t)(1+sin t) for 0 ≤ t ≤ π/2.
Now
(1 + cos t)(1 + sin t) = 1 + sin t + cos t + sin t cos t
√
= 1 + 2 sin(t + π/4) + 12 sin(2t).
The second and third terms increase from t = 0 to π/4 then decrease from t = π/4
to π/2, so the maximum occurs at t = π/4. Since the values at t = 0 and t = π/2
are equal, the minimum occurs here. This gives
»
4(1 + cos π/4)(1 + sin π/4) = 6 + 4 (2) ≈ 11.6569
as the maximum value and
4(1 + cos 0)(1 + sin 0) = 8
as the minimum value.

OC522. Find the largest natural number n such that the sum
√ √ √
b 1c + b 2c + · · · + b nc
is a prime number.
Originally from 2018 Czech-Slovakia Math Olympiad, 4th Problem, Category A,
First Round.
We received 13 submissions, of which 10 were correct and complete. We present
the solution by the Missouri State University Problem Solving Group.
Let √ √ √
f (n) = b 1c + b 2c + · · · + b nc.
Note that f (47) = 197, which is prime. We claim that this is the largest n such
that f (n) is prime.

If k 2 ≤ n ≤ (k + 1)2 − 1, then
Ñ 2
é Ñ 2
é
X−1 p
X (i+1)
k−1 n
X p k−1 X−1
X (i+1) n
X
f (n) = b jc + b jc = i + k
i=1 j=i2 j=k2 i=1 j=i2 j=k2
k−1
!
X
= (2i + 1)i + (n + 1 − k 2 )k
i=1
k(k − 1)(4k + 1)
= + (n + 1 − k 2 )k.
6

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 OLYMPIAD CORNER /381

Now f (48) = 203 = 7 · 29 is not prime.

k(k − 1)(4k + 1)
If n ≥ 49, then k > 6. Let m = ∈ Z. We claim gcd(m, k) > 1.
6
If not, then k(k − 1)(4k + 1) = 6m and gcd(m, k) = 1 would imply that k divides
6, which contradicts the fact that k > 6. If d = gcd(m, k), then d is a nontrivial
factor of f (n), since d > 1 and d ≤ k < m ≤ f (n). Therefore f (n) is not prime.
Editor’s Comment. UCLan Cyprus Problem Solving Group observed that this is
sequence A022554 in OEIS. In the comments for the sequence it is mentioned that
“It seems that 197 is the largest prime in this sequence.. . . ”. So, here it has been
confirmed that this is indeed the case.

OC523. Let (an )n≥1 be a sequence such that an > 1 and a2n+1 ≥ an an+2 for
all n ≥ 1. Prove that the sequence (xn )n≥1 defined by xn = logan an+1 for n ≥ 1
is convergent and find its limit.
Originally from 2018 Romania Math Olympiad, 3rd Problem, Grade 11, District
Round.
We received 5 solutions. We present the solution by Oliver Geupel.
Let bn = log an . By an > 1, we have bn > 0. Hence,

log an+1 bn+1

xn = = > 0.
log an bn

From the hypothesis an an+2 ≤ a2n+1 , we obtain bn + bn+2 ≤ 2bn+1 by monotoneity

1
of the logarithm. Multiplying both sides by bn+1 , we get

1
+ xn+1 ≤ 2. (1)
xn
On the other hand, it holds
1
2≤ + xn . (2)
xn
Therefore, 0 < xn+1 ≤ xn for n ≥ 1. By the monotone convergence theorem, (xn )
is convergent. Putting L = limn→∞ xn , we obtain from (1) and (2) that

1
L+ = 2;
L
whence L = 1.

OC524. Let p ≥ 2 be a natural number and let (M, ·) be a finite monoid

such that ap 6= a for all a ∈ M \ {e}, where e is the identity element of M . Prove
that (M, ·) is a group.
Originally from 2018 Romania Math Olympiad, 2nd Problem, Grade 12, District
Round.

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382/ OLYMPIAD CORNER

We received 6 solutions. We present the solution by the UCLan Cyprus Problem

Solving Group.
Take x ∈ M and consider the sequence x, x2 , x3 , . . .. Since M is finite, there are
positive integers k < ` such that xk = x` . Letting m = ` − k, by induction we can
deduce that xk = xk+mr for each r > 0. Multiplying both sides by xn−k we also
deduce that xn = xn+mr for each n > k and each r > 0.
In particular, letting a = xkm and taking n = km and r = k above, we have
a = xkm = xkm+km = a2 . By induction we now get a = at for each t > 1. In
particular a = ap and therefore a = e.
So for each x ∈ M there is an s > 1 such that xs = e. Then either s = 1 and x = e
or s > 1 and x · xs−1 = xs = e = xs = xs−1 · x. In both cases we deduce that x
has an inverse. (I.e. an element which is both a left inverse and a right inverse.)
Therefore M is a group.
Comment. Note that the fact that M is finite is needed. Otherwise (N, +) is a
counterexample.

OC525. Consider the sequence (a1 , a2 , . . . , an ) with terms from the set
{0, 1, 2}. We will call a block a subsequence of the form (ai , ai+1 , . . . , aj ), where
1 ≤ i ≤ j ≤ n, and ai = ai+1 = · · · = aj . A block is called maximal if it is not
contained in any longer block. For example, in the sequence (1, 0, 0, 0, 2, 1, 1) the
maximal blocks are (1), (0, 0, 0), (2) and (1, 1). Let Kn be the number of such
sequences of length n with terms from the set {0, 1, 2} in which all maximal blocks
have odd lengths. Moreover, let Ln be the number of all sequences of length n
with terms from the set {0, 1, 2} in which the numbers 0 and 2 do not appear in
1
adjacent positions. Prove that Ln = Kn + Kn−1 for all n > 1.
3
Originally from 2018 Poland Math Olympiad, 4th Problem, First Round.
We received 2 solutions. We present the solution by the UCLan Cyprus Problem
Solving Group.
We call a sequence nice if it is a sequence with terms from {0, 1, 2} in which all
maximal blocks have odd length. We also call a sequence good if it is a sequence
with terms from {0, 1, 2} in which the numbers 0 and 2 do not appear in adjacent
positions.
We observe that there is a 1-1 correspondence between nice sequences of length n
and nice sequences of length n + 2 whose last maximal block has length at least 3.
The correspondence is obtained by adding two elements to the sequence of length
n at its end, both equal to the last term of the sequence.
We also observe that there are 2Kn+1 nice sequences of length n + 2 whose last
maximal block has length 1. Indeed every such sequence is obtained from a nice
sequence of length n + 1 by adding at its end a term which is different from the
last term of the sequence.

Crux Mathematicorum, Vol. 47(8), October 2021

 OLYMPIAD CORNER /383

Thus Kn+2 = 2Kn+1 + Kn for each n > 1. Defining K0 = 0, then the condition
holds for n = 0 as well.
Now let An , Bn , Cn be the number of good sequences of length n whose last term
is equal to 0, 1, 2 respectively. By symmetry we have An = Cn . Then An+1 =
An + Bn and Bn+1 = An + Bn + Cn = 2An + Bn for each n > 1. Defining
A0 = C0 = 0 and B0 = 1, then the condition holds for n = 0 as well. Then

Ln+2 = 2An+2 + Bn+2 = 4An+1 + 3Bn+1 = 2Ln+1 + Bn+1 = 2Ln+1 + Ln

for each n > 0.

Defining Mn = Kn + 31 Kn−1 for n > 1, we observe that Mn satisfies the same
recurrence relation as Ln since

6Mn+1 + 3Mn = (6Kn+1 + 2Kn ) + (3Kn + Kn−1 )

= 6Kn+1 + 5Kn + (Kn+1 − 2Kn )
= (6Kn+1 + 3Kn ) + Kn+1
= 3Kn+2 + Kn+1 = 3Mn+2

We now observe that L1 = M1 = 3 and L2 = M2 = 7, so Ln = Mn = Kn + 31 Kn−1

for each n > 1.

Copyright © Canadian Mathematical Society, 2021

384/ An introduction of the problem of finding the chromatic number of the plane (II)

An introduction of the problem of

finding the chromatic number of
the plane (II)
Veselin Jungić

1 Introduction

In the first part of this two-part note [5], we introduced the chromatic number of
the plane problem:
Problem 1 What is χ, the smallest number of sets (“colours”) with which we
can cover the plane in such a way that no two points of the same set are a unit
distance apart? [1]
Through a series of examples and exercises, we guided the reader to establish that
4 ≤ χ ≤ 7. In particular, we used the unit distance graph called the Moser spindle,
Figure 1, to show that χ ≥ 4 since 3 colours are not enough for a proper colouring
that avoids adjacent vertices of the same colour. The Moser spindle was originally
introduced by Canadian mathematicians Leo and William Moser in 1961. [6]

Figure 1: The Moser spindle.

Regardless of all efforts, until very recently, the lower bound for χ, established by
an American mathematician Edward Nelson in the 1950s [12] remained unchanged.

2 de Grey’s theorem

In 1979, the renowned Hungarian mathematician Paul Erdős wrote:

It seems likely that the chromatic number [of the plane] is greater than
four. By a theorem of de Bruijn and myself this would imply that there
are n points x1 , x2 , . . . , xn in the plane so that if we join any two of
them whose distance is 1, then the resulting graph G(x1 , x2 , . . . , xn )

Crux Mathematicorum, Vol. 47(8), October 2021

 Veselin Jungić /385

has a chromatic number > 4. I believe such an n exists but its value
may be very large. [3, p.156]
Almost forty years later, in 2018, an English author and biomedical gerontologist
Aubrey de Grey proved that Erdős was correct on both accounts. de Grey con-
structed a unit distance graph with 20425 vertices and with its chromatic number
greater than 4.
Therefore, de Grey proved the following:
Theorem 1 χ ≥ 5. [2]
de Grey’s proof was a combination of new insights into some of the well-known
facts and techniques and a computer assisted mathematical proof. In de Grey’s
words:
In seeking graphs that can serve as [a unit distance graph with the
chromatic number greater than four] in our construction, we focus on
graphs that contain a high density of Moser spindles. The motivation
for exploring
√ such graphs is that a spindle contains two pairs of vertices
distance 3 apart, and these pairs cannot both be monochromatic. In-
tuitively, therefore, a graph containing a high density of interlocking
√
spindles might be constrained to have its monochromatic 3-apart
vertex pairs distributed rather uniformly (in some sense) in any 4-
colouring. Since such graphs typically also contain regular hexagons
of side-length 1, one might be optimistic that they could contain some
such hexagon that does not contain a monochromatic triple in any
4-colouring of the overall
√ graph, since such a triple is always an equi-
lateral triangle of edge 3 and thus constitutes a locally high density,
i.e.
√ a departure from the aforementioned uniformity, of monochromatic
3-apart vertex pairs. [2]
Initially, de Grey “developed a custom program” to test his original graph on
20425 vertices for the existence of monochromatic points with the unit distance
under a 4-colouring. de Grey explained that “this algorithm was implemented
in Mathematica 11 on a standard MacBook Air and terminated in only a few
minutes.”
In the following exercises, we will explore some terms from the above quote and
construct three “tightly linked Moser spindles,” one of the tools that de Grey used
in his proof.
Exercise 1 Use a graphing tool, Desmos or GeoGebra, for example, to construct
the Moser spindle with equilateral triangles of side length 1. Let A be the vertex
of degree four and let C and D be two vertices that are not adjacent to A. Let α
be the measure of ∠DAC in degrees. See Figure 2.
i) Determine |AC| and |AD|.
ii) Determine the value of α.

Copyright © Canadian Mathematical Society, 2021

386/ An introduction of the problem of finding the chromatic number of the plane (II)

C
D

Figure 2: Moser spindle and the angle α

iii) Construct the Moser spindle by starting with drawing two concentric circles,
one with radius 1 and the other with radius |AC|.
Exercise 2 Confirm de Grey’s observation
√ from the quote above that “a spindle
contains two pairs of vertices distance 3 apart, and these pairs cannot both be
monochromatic” in a proper k-colouring of the Moser spindle, k > 3.
Exercise 3 Use your graphing tool to rotate clockwise the Moser spindle through
α about the point A to obtain another Moser spindle. Call the newly obtained
graph, i.e. the graph that consists of the original Moser spindle and its image
obtained by the rotation, M S2 . What do you observe? Is this a unit distance
graph? Does M S2 contain a hexagon, not necessarily regular, “of side-length 1?”
Exercise 4 Find a proper 4-colouring of M S2 . Do this in two ways, by hand and
by writing a program that would search for proper 4-colourings of M S2 .
Exercise 5 Alter the drawing of M S2 by rotating clockwise the original Moser
spindle by α2 about the point A. Find a proper 4-colouring of the newly obtained
graph.
We would like to encourage the reader to explore other graphs, tools, and ideas
that de Grey presents in his remarkable paper.
Already in his original paper, de Grey was able to reduce the size of the unit
distance graph with the chromatic number greater than four to 1581. Again, in
de Grey’s words, “Happily, G has turned out to be within the reach of standard
SAT solvers with which others have now confirmed its chromatic number to be 5
without the need to resort to using custom code or checking weaker properties of
subgraphs.”
On August 3, 2019, as part of the Polymath 16 project [9], Jaan Parts from Kazan,
Russia, posted an image of a unit distance graph with 510 vertices and 2508 edges
that confirms that χ ≥ 5.

Crux Mathematicorum, Vol. 47(8), October 2021

 Veselin Jungić /387

3 Conclusion
The goals of the researchers involved in the Polymath 16 project included:
1. To further reduce the size of the “good” graph;
2. To find a human-verifiable proof that χ ≥ 5.
Clearly, the two goals are interrelated. A graph of relatively small size found by a
computer may lead to “a human-verifiable proof that χ ≥ 5.” And the other way
around, a moment of human brilliance may provide a proof that would contain,
if not the exact value, then a reasonable bound for the size of the minimal unit
distance graph with a chromatic number greater than four.
It seems that the first scenario is more likely. For example, in their recent paper
that includes some of the authors’ contributions to the Polymath 16 project, Nóra
Frankl, Tamaás Hubai, and Dömötör Pálvölgyi wrote:
Note that G34 [a graph on 34 vertices that cannot be 4-coloured with
the so-called “bichromatic origin”] was found by a computer search,
and for finding other similar graphs one might rely on a computer
program. Thus, the approach we propose, is human-verifiable, however
it might be computer-assisted. [4]
In 2020, Parts contributed to both of the above listed goals. In [7], Parts described
a set of ideas that led to a construction of a 5-chromatic graph with 509 vertices
and 2442 edges. Parts also came up with a human-verifiable proof that χ ≥ 5:
The proof presented here may leave a sediment of dissatisfaction. We
tend to explain this by the fact that we expect not only a human-
verifiable proof, but a really human proof. The difference is that in
the latter case, the proof is created by a person, while in the first case
it is not necessary: it is only important that the person is able to
complete the verification in a reasonable number of steps. [8]
The recent developments in the search for the chromatic number of the plane have
further highlighted the following two realities:
1. Mathematical research has become increasingly collaborative;
2. Computing has become an instrumental part of mathematical research.
The Polymath 16 project is winding down. A paper that will summarize the
outcomes of the project is expected in the near future. In a post dated February
1, 2021, de Grey concluded:
Perhaps the most satisfying aspect of the project, to me, is its quite
remarkable success in attracting amateur mathematicians like myself
(of whom Jaan is one). At least half a dozen of us have been significant
contributors over these three years, working side by side with numerous
professionals including Terry Tao himself. [10]
Still, is χ = 5? Or maybe it is χ = 6? Or, perhaps, χ = 7?

Copyright © Canadian Mathematical Society, 2021

388/ An introduction of the problem of finding the chromatic number of the plane (II)

4 Hints and solutions

√ 5
≈ 33.560 . iii) see below.


Exercise 1. i) |AC| = |AD| = 3. ii) α = arccos 6

Step 1
C
Start by choosing a point A in
the plane and then draw a circle
with the centre at A and radius
1. Denote this circle by c1 . Next,
draw a circle with
√ the centre at
A A and radius 3. Denote this
circle by c2 . Choose a point C
on the circle c2 .

Step 2 Draw a circle with the centre at

D C C and radius 1. Denote this cir-
cle by c3 . Let D be a point of
intersection of c2 and c3 . Let
F B and F be the points of in-
B
tersection of c1 and c3 . Draw
the line segments AB, AF , BC,
A BF , CD, and CF . Observe that
those line segments are of length
1. How would you argue that
|BF | = 1?

Step 3
D C Draw a circle with the centre at
D and radius 1. Denote this cir-
cle by c4 . Let E and G be the
F G points of intersection of c1 and
B
E c4 . Draw the line segments AE,
AG, DE, DG, and EG. Observe
A that those line segments are of
length 1. The Moser spindle ap-
pears!

Exercise 2. If both (A, C) and (A, D) are monochromatic then the pair (C, D)
is monochromatic as well.

Crux Mathematicorum, Vol. 47(8), October 2021

 Veselin Jungić /389

Exercise 3. Observe that the two spindles share four vertices and five edges.
This is what de Grey meant when saying “interlocking spindles.” Also note that
the outer polygon is a hexagon of side-length 1.
C

Exercise 4. Say that A is the vertex of degree 6. Colour A blue and then colour
the vertices not adjacent to A by yellow and green alternatively. Use blue and red
to colour the remaining vertices. To write a program you may use the following
strategy. Denote the four colours by the numbers 1, 2, 3, and 4 . Use 0 to
denote the vertex that is not coloured. Then go through each vertex that was not
coloured, get all the numbers of the vertices around it, then colour the vertex by
the lowest positive number that is not in that set. For example, if a vertex had
vertices around it with the numbers 0, 1, 2, and 2, then this vertex would have to
be given the number 3 (which means colour number three).

α
Exercise 5. An additional clockwise rotation by 2 produces, what de Grey
calls,“three tightly linked Moser spindles.”

Use the same idea as in Exercise 4 to find a proper 4-colouring by hand. Al-
ternatively, you may write a program that generates proper 4-colourings. Here

Copyright © Canadian Mathematical Society, 2021

390/ An introduction of the problem of finding the chromatic number of the plane (II)

is a 4-colouring generated by a Python program written by Ewan Brinkman, a

first-year student at Simon Fraser University:

√
Do you notice what de Grey calls “monochromatic 3-apart vertex pairs (. . . ) in
any 4-colouring”?

References

[1] Croft, H.T., Falconer, K.J., and Guy, R.K. Unsolved Problems in Geometry.
Problem Books in Mathematics. Springer-Verlag, New York, 1991. Unsolved
Problems in Intuitive Mathematics, II.
[2] de Grey, Aubrey D.N.J. (2018), The chromatic number of the plane is at least
5. Geombinatorics, 28, p. 5–18. arXiv:1804.02385.
[3] Erdős, P., Combinatorial problems in geometry and number theory. Proc. of
the Sympos. Pure Math., 34, Amer. Math. Soc., Providence, RI, (1979), p.
149–162.
[4] Frankl, N., Hubai, T., and Pálvölgyi, D. Almost-monochromatic sets and the
chromatic number of the plane. Proceedings of SoCG 2020. arXiv:1912.02604.
[5] Jungić V., An introduction of the problem of finding the chromatic number of
the plane (I), Crux Mathematicorum, 46 (8), p. 390–396.
[6] Moser, L., and Moser, W., Solution to Problem 10, Can. Math. Bull. 4 (1961),
p. 187–189.
[7] Parts, Jaan. (2020), Graph minimization, focusing on the example of 5-
chromatic unit-distance graphs in the plane, Geombinatorics, 29, p. 137–166.
arXiv:2010.12665.
[8] Parts, Jaan. (2020), The chromatic number of the plane is at least 5 – a human-
verifiable proof, Geombinatorics, 30, p. 11–102. arXiv:2010.12661.

Crux Mathematicorum, Vol. 47(8), October 2021

 Veselin Jungić /391

[9] Polymath16 Project. https://dustingmixon.wordpress.com/2019/07/08/polymath16-

thirteenth-thread-bumping-the-deadline/#comment-23999
[10] Polymath16 seventeenth thread: Declaring victory.
https://dustingmixon.wordpress.com/2021/02/01/polymath16-seventeenth-
thread-declaring-victory
[11] Soifer, A. The mathematical coloring book, Springer, New York, 2009
[12] Soifer, A. Chromatic number of the plane & its relatives, history, problems
and results: an essay in 11 parts, Ramsey theory, Progr. Math., vol. 285,
Birkhäuser/Springer, New York, 2011, p. 121–161.

Copyright © Canadian Mathematical Society, 2021

392/ 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 December 30, 2021.

4671. Proposed by Minh Ha Nguyen.

Given a triangle ABC and a point M on the side BC, let r1 and r2 be inradii
of triangles ABM and ACM , respectively. Determine the position of M which
yields the maximum value of r1 + r2 .

4672. Proposed by Nguyen Viet Hung.

Let a, b, c be rational numbers such that
p p p
a2 − ab + b2 + b2 − bc + c2 = a2 + ac + c2 .
√
Prove that a2 + b2 + c2 is a rational number.

4673. Proposed by Todor Zaharinov.

Let P be a point in the plane of triangle ABC not on any line joining two of its
vertices. Let P A1 A2 be the triangle with centroid A and vertices A1 ∈ BA and
A2 ∈ AC. By cyclically permuting A, B, C we similarly define triangle ∆P B1 B2
with centroid B, and P C1 C2 with centroid C. Prove that the triangles A1 B1 C1
and A2 B2 C2 have equal areas.

4674. Proposed by Michel Bataille.

Find all monic polynomials P (x) ∈ Z[x], of positive degree 2n, that have a complex
root of multiplicity n and satisfy P (0) = −1 and P (1) · P (−1) 6= 0.

Crux Mathematicorum, Vol. 47(8), October 2021

 Problems /393

4675. Proposed by George Apostolopoulos.

Let a, b and c be positive real numbers such that a2 + b2 + c2 = 3. Prove that

2(a4 + b4 + c4 ) − (a3 + b3 + c3 ) ≥ 3abc.

4676. Proposed by Lorian Saceanu.

Let ABCD be a convex quadrilateral with E at the intersection of diagonals.
From E build the bisectors of the angles ∠AEB, ∠BEC, ∠CED, ∠DEA. We
get an orthodiagonal quadrilateral F GHI with F ∈ [AB], G ∈ [BC], H ∈ [CD],
I ∈ [DA]. Prove that:
[AF ] [BG] [CH] [DI]
· · · = 1.
[BF ] [CG] [DH] [AI]

4677. Proposed by Seán M. Stewart.

Evaluate ∞
2x
Z Å ã
x
arctan dx.
0 1 + x2 x2 +4

4678. Proposed by Michel Bataille.

Let n be a nonnegative integer. Evaluate in closed form
bn/2c Ç å Å ãj
X n+1 1
j=0
n − 2j 5
bn/2c ã .
1 j
Ç å
n−j
X Å
−
j=0
n − 2j 5

4679. Proposed by Daniel Sitaru.

Let (xn )n≥1 be a sequence of real numbers such that x1 = 71 , x2 = 1
5 and for n ≥ 2,

2nxn+1 · xn−1 = (n + 1)xn · xn−1 + (n − 1)xn · xn+1 .

Find 2 nxn
lim + xn .
n→∞ 3

4680. Proposed by Pericles Papadopoulos.

In triangle ABC with |BC| > |AC| > |AB|, the circle C1 , centered at C and
with radius CA, meets the sides AB, BC at points P, Q, respectively and the

Copyright © Canadian Mathematical Society, 2021

394/ Problems

circle C2 , centered at B and with radius BA, meets the sides AC, BC at points
S, T , respectively. Let A0 be the second intersection point of C1 and C2 and let
O1 , O2 , O3 , O4 , O5 be the circumcenters of triangles CST, SAP, P BQ, BA0 Q, A0 CT ,
respectively. Prove that the points C, O1 , S, O2 , P, O3 , B, O4 , A0 , O5 are concyclic.

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

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 30 decembre 2021.

4671. Proposé par Minh Ha Nguyen.

Soit un triangle ABC et M un point sur le côté BC, puis soient r1 et r2 les rayons
des cercles inscrits des triangles ABM et ACM respectivement. Déterminer où
situer M afin d’obtenir la valeur maximale de r1 + r2 .

4672. Proposé par Nguyen Viet Hung.

Soient a, b, c des nombres rationnels tels que
p p p
a2 − ab + b2 + b2 − bc + c2 = a2 + ac + c2 .
√
Démontrer que a2 + b2 + c2 est un nombre rationnel.

Crux Mathematicorum, Vol. 47(8), October 2021

 Problems /395

4673. Proposé par Todor Zaharinov.

Soit P un point dans le plan du triangle ABC, mais ne se trouvant pas sur les
lignes rejoignant deux de ses sommets. Soit P A1 A2 le triangle de centroı̈de A
et sommets A1 ∈ BA et A2 ∈ AC. Par permutation cyclique, on définit les
triangles P B1 B2 et P C1 C2 de centroı̈des B et C respectivement. Démontrer que
les triangles A1 B1 C1 et A2 B2 C2 ont la même surface.

4674. Proposé par Michel Bataille.

Déterminer tout polynôme unitaire P (x) ∈ Z[x], de degré positif 2n, tel que
P (1) · P (−1) 6= 0, P (0) = −1 et ayant une racine complexe de multiplicité n.

4675. Proposé par George Apostolopoulos.

Soient a, b et c des nombres réels positifs tels que a2 + b2 + c2 = 3. Démontrer que

2(a4 + b4 + c4 ) − (a3 + b3 + c3 ) ≥ 3abc.

4676. Proposé par Lorian Saceanu.

Soit ABCD un quadrilatère convexe et soit E le point d’intersection de ses di-
agonales. À partir de E, construire les bissectrices des angles ∠AEB, ∠BEC,
∠CED, ∠DEA. Il en suit un quadrilatère à diagonales orthogonales F GHI, où
F ∈ [AB], G ∈ [BC], H ∈ [CD], I ∈ [DA]. Démontrer que

[AF ] [BG] [CH] [DI]

· · · = 1.
[BF ] [CG] [DH] [AI]

4677. Proposé par Seán M. Stewart.

Évaluer
∞
2x
Z Å ã
x
arctan dx.
0 1 + x2 x2 +4

Copyright © Canadian Mathematical Society, 2021

396/ Problems

4678. Proposé par Michel Bataille.

Soit n un entier non négatif. Évaluer en forme close
bn/2c
n+1 1 j
P


n−2j 5
j=0
.
bn/2c
j
n−j
− 51
P

n−2j
j=0

4679. Proposé par Daniel Sitaru.

Soit (xn )n≥1 une suite de nombres réels telle que x1 = 17 , x2 = 1
5 et, pour n ≥ 2,

2nxn+1 · xn−1 = (n + 1)xn · xn−1 + (n − 1)xn · xn+1 .

Déterminer 2 nxn
lim + xn .
n→∞ 3

4680. Proposé par Pericles Papadopoulos.

Soit ABC un triangle où |BC| > |AC| > |AB|. Supposons que le cercle C1 , centré
à C et de rayon CA, rencontre les côtés AB et BC en P et Q respectivement, puis
que le cercle C2 , centré à B et de rayon BA, rencontre les côtés AC et BC en S
et T respectivement. Soit A0 le second point d’intersection de C1 et C2 et soient
O1 , O2 , O3 , O4 , O5 les centres des cercles circonscrits de CST , SAP , P BQ, BA0 Q,
A0 CT , respectivement. Démontrer que les points C, O1 , S, O2 , P , O3 , B, O4 , A0 ,
O5 sont cocycliques.

Crux Mathematicorum, Vol. 47(8), October 2021

 Solutions /397

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 2021: 47(3), p. 151–154.

4621. Proposed by Michel Bataille.

In the plane, circles C and C 0 are externally tangent at T . Points P, Q of C 0 are
such that P 6= T and ∠P QO = 90◦ where O is the centre of C. Points A, B of C
are such that P A and QB are tangent to C. If the line P T intersects C again at
S, prove that P S · QT = P A · QB.
We received 7 submissions, all correct, and present the solution by Borche Jo-
shevski.

The requirement that ∠P QO = 90◦ is not needed; instead, we let P and Q be

arbitrary points of the circle C 0 with P distinct from Q and T . From the figure on
the left, we see that ∆SOT = ∆P O0 T (where O and O0 are the centers of C and
C 0 ). It follows that
PT O0 T r0
= = . (1)
ST OT r
If Q and T coincide, then QT = 0 = QB, and the condition P S · QT = P A · QB
is satisfied trivially. So we take Q 6= T and define R to be the point where QT
0
again intersects C. Equation (1) implies that PSTT = rr = QTRT , or

PT TS PT + TS PS
= = = .
QT TR QT + T R QR
That is,
P T · QR = P S · QT. (2)
2
The powers of P and Q with respect to C satisfy both P T · P S = P A and
QT · QR = QB 2 . By multiplying these two equalities and applying equation (2)

Copyright © Canadian Mathematical Society, 2021

398/ Solutions

we get, in turn,

P T · P S · QT · QR = P A2 · QB 2 ,
(P T · QR) · (P S · QT ) = (P A · QB)2 ,
(P S · QT )2 = (P A · QB)2 ,
P S · QT = P A · QB.

4622. Proposed by Mihaela Berindeanu.

Let ABC be an acute triangle and AA0 , BB 0 and CC 0 be its medians. Let Γ
be the circumcircle of M A0 B 0 C 0 and let Γ ∩ AA0 = {A00 }, Γ ∩ BB 0 = {B 00 },
−−→ −−−→ −−→ →
−
Γ ∩ CC 0 = {C 00 }. Show that if AA00 + BB 00 + CC 00 = 0 , then ABC is an
equilateral triangle.
We received 9 submissions, but only five were complete and correct. We present
the solution by Theo Koupelis.
Because A0 , B 0 , C 0 are the midpoints of the sides of the triangle, the circle Γ is
its Euler circle. As such, the feet of the altitudes also belong on Γ, and because
the triangle is acute, these points are on the sides AB, BC, CA and not on their
extensions; therefore, Γ cuts the medians at points A00 , B 00 , C 00 that are inside the
triangle.
Let a, b, c be the lengths of the triangle’s sides, let G be the centroid, and let D
be the foot of the altitude from C to AB. Then the power of point A with respect
to Γ is equal to AD · AC 0 = AA00 · AA0 . But

b2 + c2 − a2 c
AD = b · cos A = , and AC 0 = .
2c 2
1 9
Therefore, with AA0 = ma , where m2a = · (2b2 + 2c2 − a2 ) = · AG2 , we get
4 4
b2 + c2 − a2
AA00 = , and therefore
4ma
−→
−−→ AG 3(b2 + c2 − a2 ) −→ −→ a2 + b2 + c2 −→
AA00 = AA00 · = · AG = AG − · AG.
AG 8m2a 8m2a
−−−→ −−→
Adding similar expressions for BB 00 , CC 00 , and using the well-known expression
−→ −−→ −−→ → −
AG + BG + CG = 0 , we get
→
− −−→ −−−→ −−→
0 = AA00 + BB 00 + CC 00
Ç −→ −−→ −−→ å
−→ −−→ −−→ a2 + b2 + c2 AG BG CG
= (AG + BG + CG) − · + 2 + 2
8 m2a mb mc
ma 2 ma 2
®ñÅ ô ñ ô ´
a2 + b2 + c2 −−→ −−→
ã Å ã
= − · − 1 · BG + − 1 · CG .
8m2a mb mc

Crux Mathematicorum, Vol. 47(8), October 2021

 Solutions /399

−−→ −−→
But BG and CG are linearly independent vectors on the plane of the triangle and
therefore we must have ma = mb = mc , and thus a = b = c and the triangle is
equilateral.

Editor’s comments. Michel Bataille observed that it is not necessary to suppose

that ∆ABC is acute. He showed that the proposer’s result is equivalent to
Theorem. Suppose triangle ABC, with centroid G, is inscribed in the circle Ω
while the medians from A, B, C intersect Ω again at A1 , B1 , C1 . Then the relation
−−→ −−→ −−→ → −
GA1 + GB1 + GC1 = 0 implies that ∆ABC is equilateral.
−→ −−→ −−→ → −
He argued as follows: If G is the centroid of ∆ABC, then GA + GB + GC = 0 .
−−→ −−−→ − −→ →
− −−→ −−→ −−→ →
−
It follows that AA00 + BB 00 + CC 00 = 0 is equivalent to GA00 + GB 00 + GC 00 = 0 .
0 0 0
Moreover, G is also the centroid of ∆A B C , and ∆ABC is equilateral if and only
if ∆A0 B 0 C 0 is. In other words, by applying Bataille’s result to prove that ∆A0 B 0 C 0
is equilateral, it follows that the given triangle ABC is likewise equilateral.

4623. Proposed by Nguyen Viet Hung.

Let p(x) = x2 + 2ax − b − 1 and q(x) = x2 + 2bx − a − 4 be two polynomials with
integer coefficients. Determine all pairs (a, b) of non-negative integers such that
these two polynomials simultaneously have integer solutions.
We received 13 correct and 3 incorrect solutions. We present the solution of Cor-
neliu Avram-Manescu and the proposer (done independently).
Since

p(x) = (x + a)2 − (a2 + b + 1) and q(x) = (x + b)2 − (b2 + a + 4)

have integer roots, both a2 + b + 1 and b2 + a + 4 are integer squares exceeding a2

and b2 respectively. Since

a2 + b + 1 ≥ (a + 1)2 and b2 + a + 4 ≥ (b + 1)2 ,

it follows that b ≥ 2a and a + 3 ≥ 2b ≥ 4a. Therefore a = 0 or a = 1.

If a = 0, then b2 + 4 is a square, which implies that b = 0. In this case,

p(x) = x2 − 1 = (x − 1)(x + 1)

and
q(x) = x2 − 4 = (x − 2)(x + 2).

If a = 1, then b2 + 5 is a square, which implies that b = 2. In this case,

p(x) = x2 + 2x − 3 = (x − 1)(x + 3)

and
q(x) = x2 + 4x − 5 = (x − 1)(x + 5).

Copyright © Canadian Mathematical Society, 2021

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4624. Proposed by George Apostolopoulos.

Let a, b, c be positive real numbers with a + b + c = 3. Prove that
   
3
…
ab bc ca
+ + ≤ .
2a + b + c 2b + c + a 2c + a + b 2

We received 24 submissions of which 23 were correct and complete. Most solutions

made explicit use of one or more of Cauchy’s inequality, Jensen’s inequality, and
the AM-HM inequality. We present two solutions.
Solution 1, by Antonio Garcia and Nikos Ntorvas (done independently).
Let S denote the left-hand side of the inequality. Since a + b + c = 3, and by
Cauchy’s inequality we have
      !
√
… … …
ab bc ca a b c
S= + + ≤ b+c+a + +
3+a 3+b 3+c 3+a 3+b 3+c
 
√ a b c
= 3· + + .
3+a 3+b 3+c

Hence, it suffices to show that

a b c 3
+ + ≤ .
3+a 3+b 3+c 4
x
Next, define f (x) = 3+x for x > 0 so that f 00 (x) = −6(3 + x)−3 < 0. Hence, f is
concave on (0, ∞) and by Jensen’s inequality

a+b+c 3
f (a) + f (b) + f (c) ≤ 3 · f ( ) = 3 · f (1) = .
3 4

Solution 2, by the proposer and numerous others.

By the HM-AM inequality, we have the following cyclic sums:

1X
Å ã
X a X a a a
= ≤ +
2a + b + c (a + b) + (a + c) 4 a+b a+c
1X 3
Å ã
a b
= + = .
4 a+b b+a 4
»
ab
Next, define S = 2a+b+c . So, by Cauchy’s inequality, and since a + b + c = 3,
P

X a 3
S 2 ≤ (b + c + a) ≤3· .
2a + b + c 4

Hence, S ≤ 23 . In fact, the conclusion holds under the hypothesis that a+b+c ≤ 3.

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4625. Proposed by Corneliu Manescu-Avram.

Let a, m and n be positive integers greater than 1. Prove that a2 + a + 1 divides
(a + 1)m + an if and only if m is odd and 3 divides m + n.
We received 18 submissions, 4 of which were incomplete having only proven one
direction of the “if and only if” statement. We present the solution by the editorial
board.
Let r = a2 + a + 1. Then modulo r, we have a2 ≡ −(a + 1), a3 ≡ 1 and ak ≡ at ,
where k ≡ t (mod 3) for some t with 0 ≤ t < 3.
If m is even, then (a + 1)m + an ≡ a2m + an (mod r). Modulo r, a2m + an is of
the form au (1 + av ), where u, v ∈ Z such that 0 ≤ u, v < 3. Since 0 < 1 + av < r,
r does not divide 1 + av , so the divisibility is never possible when m is even.
If m is odd, then (a + 1)m + an ≡ −a2m + an (mod r), which is divisible by r if
and only if am (−a2m + an ) = −a3m + am+n ≡ −1 + am+n (mod r) is divisible by
r. But am+1 ≡ 1 (mod r) if and only if 3|(m + n), so r|((a + 1)m + an ) if and only
if m is odd and 3|(m + n), which completes the proof.

4626. Proposed by Alpaslan Ceran.

Let ABCDE and ABF KM be regular pentagons and suppose that AC intersects
EB at a point L. Show that
√
|LK| 5+1
= .
|DL| 2

We received 22 solutions, all of which were correct. We present the solution by

Theo Koupelis.

Let a be the side length of the regular pentagon whose vertex angle is 108◦ . Tri-
angles ABC and EAB are congruent and thus their base angles are equal to 36◦ .

Copyright © Canadian Mathematical Society, 2021

402/ Solutions

Therefore ∠EAC = ∠BED = 108◦ −36◦ = 72◦ and thus triangle EAL is isosceles,
with ∠ELA = 72◦ and EL = a. As a result, triangle LED is also isosceles with
base angles equal to 54◦ . Therefore, if M is the midpoint of side AB, the points
M, L, D are collinear and DM is the perpendicular bisector of AB. Similarly, be-
cause ABCDE and ABF KN are congruent, KM is the perpendicular bisector of
AB, and thus the points K, M, L, D are collinear and KM = M D.
Triangles ALB and EAB are similar and thus AB 2 = BL √
· BE or a2 = BL ·
5−1
(BL + a). Solving the quadratic in BL we find BL = a · 2 . From the right
√
triangle BM L we have a = 2 · BL · cos 36◦ , and therefore cos 36◦ = 5+1 4 . Also,
LM = a2 · tan 36◦ , and DL = 2a · cos 54◦ = 2a sin 36◦ . Thus,
√
|LK| |LM | + |DM | |LM | 1 5+1
= =1+2· =1+ = .
|DL| |DL| |DL| 2 cos 36◦ 2

4627. Proposed by Dong Luu.

Let P be a point not on the circumcircle (O) of a given triangle ABC, nor on
the extensions of any of its sides. Define U, V, W to be the projections of P
on the lines BC, CA, AB, and X, Y, Z to be the vertices of the triangle formed
by the perpendicular bisectors of the segments P A, P B, P C. Suppose that the
circumcircle of ∆XY Z intersects (O) in two points, E and F . Prove that the foot
of the perpendicular from P to the line EF is the circumcenter of triangle U V W .
The only submission came from the proposer himself. As you can see from his
solution (which was only slightly modified by the editor), the problem is rather
interesting, and so it is a mystery to this editor why it failed to attract more
attention.
Recall that two lines through the vertex of a directed angle are isogonal with
respect to that angle if they are symmetrical about the angle bisector; in other
words, they share the same angle bisector.

Lemma. Given two lines, ` and m that intersect in a point O, and a second point
P in their plane, define L and M to be the projections of P on ` and m. Then if

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 Solutions /403

P 0 is an arbitrary point on the line isogonal to OP (with respect to ` and m), the
midpoint S of P P 0 is equidistant from L and M .

Proof of the lemma. Because of the right angles at L and M , OP is the diameter
of the circumcircle of triangle OLM . It is known (and easier to prove than to find
in a reference) that the line OP (that joins the vertex O to the circumcenter, call
it O0 ) and the altitude to LM from O are isogonal with respect to the sides OL
and OM ; that is, OP 0 is perpendicular to the side LM . But the line joining O0
(the circumcenter and, therefore, the midpoint of OP ) to S (the midpoint of P P 0 )
is parallel to OP 0 , and therefore is the perpendicular bisector of the chord LM of
the circumcircle. Consequently, S is equidistant from L and M , as claimed.

Proof of the main result. Define P 0 to be the isogonal conjugate of P with respect
to ∆ABC. (That is, the lines joining P and P 0 to any vertex are isogonal with
respect to the angle at that vertex.) The lemma implies that the midpoint of
P P 0 is equidistant from the points U, V , and W , and is thus the circumcenter of
∆U V W . The problem is thereby reduced to proving that EF is the perpendicular
bisector of P P 0 . To that end we define a new triangle X 0 Y 0 Z 0 whose vertices are
the intersections of the perpendicular bisectors of P 0 A, P 0 B, and P 0 C. Because X
is on the perpendicular bisectors of P B and P C, we deduce that X is equidistant
from B, P , and C. Similarly, X 0 is equidistant from those same points. It follows
that X and X 0 are both on the perpendicular bisector of BC, whence O, X, X 0

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404/ Solutions

are collinear.
We now use directed angles to prove that OB is tangent to the circle (BXX 0 ):

1
∠XBO = ∠BXO + ∠XOB = (∠BXC + ∠COB)
2
= ∠BP C + CAB
= ∠P BC + ∠BCP + ∠CBA + ∠ACB
= (∠CBA + ∠P BC) + (∠ACB + ∠BCP )
= ∠P BA + ∠ACP
= ∠CBP 0 + ∠P 0 CB
1
= ∠CP 0 B = ∠CX 0 B
2
= ∠OX 0 B = ∠XX 0 B,

which implies that OB is tangent to (BXX 0 ), as claimed. As a consequence,

this circle is perpendicular to the circumcircle (O) of the given ∆ABC, whence
inversion in (O) interchanges X with X 0 . Similarly, this inversion interchanges Y
with Y 0 and Z with Z 0 while it fixes the points E and F (which are the points
common to the circles (O), (XY Z), and therefore also to (X 0 Y 0 Z 0 ). In other
words, the line EF contains the points whose power with respect to (XY Z) equals
their power with respect to (X 0 Y 0 Z 0 ). Furthermore, X, X 0 , Y, Y 0 are concyclic
points (because any circle through a point and its inverse is perpendicular to the
circle of inversion, and is therefore fixed by the inversion; that is, the line OY
must meet the circle (XX 0 Y ) again at Y 0 ). This implies that the lines XY and
X 0 Y 0 must intersect at a point, call it K on the line EF (because the powers
KX · KX 0 = KY · KY 0 , so that K must lie on the line of equal powers with
respect to the circles (XY Z) and (X 0 Y 0 Z 0 ). Recall that XY is the perpendicular
bisector of P C, while X 0 Y 0 is the perpendicular bisector of P 0 C 0 ; consequently,
K is equidistant from P, C, and P 0 . Similarly, XZ intersects X 0 Z 0 at a point
of EF (distinct from K) that is equidistant from P and P 0 . We conclude that
EF is the perpendicular bisector of P P 0 ; since S is (by definition) the foot of
the perpendicular from P to EF , it must be the midpoint of P P 0 , which (by the
lemma) completes the proof.

4628. Proposed by Russ Gordon and George Stoica.

Let A be a nonempty set of positive integers that is closed under addition and such
that N \ A contains infinitely many elements. Prove that there exists a positive
integer d ≥ 2 such that A ⊆ {nd : n ∈ N}.
We received 7 submissions and they were all complete and correct. We present two
solutions.
Solution 1, by Oliver Geupel, Roy Barbara and UCLan Cyprus Problem Solving
Group (done independently), slightly modified by the editor.

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 Solutions /405

Let a ∈ A. Note that if a = 1, then A = N, contradicting to our assumption.

Thus, a > 1. Let p1 , . . . , pk be the prime divisors of A. If for some 1 6 i 6 k we
have A ⊆ {npi : n ∈ N} then we are done.
So we may assume that for each 1 6 i 6 k there is an ai ∈ A with pi - ai .
Then we have gcd(a, a1 , . . . , ak ) = 1. This would imply that each positive integer
n > F (a, a1 , . . . , ak ) can be written as an integer linear combination of a,a1 ,. . .,ak
with non-negative coefficients, where F (a, a1 , . . . , ak ) is the Frobenius number as-
sociated to a,a1 ,. . ., ak .
Note that gcd(a, a1 , . . . , ak ) = 1 implies that F (a, a1 , . . . , ak ) < ∞. In other words,
N \ A is finite, a contradiction.

Solution 2, by the proposer, slightly modified by the editor.

Fix some k ∈ A. We have that k ≥ 2, otherwise A = N, a contradiction.
In the following discussion, we would identify i ∈ Zk with its smallest positive
residue modulo k. For each i ∈ Zk , we denote

[i] := {i, i + k, i + 2k, i + 3k, . . .}, Ai = [i] ∩ A.

By the additivity assumption, if Ai 6= ∅, then [i] \ Ai is finite.

Let G = {i ∈ Zk : Ai 6= ∅}. We claim that G is a subgroup of Zk . Indeed, if
i, j ∈ G, then i + j ∈ G and −i = (k − 1)i ∈ G since A is closed under addition.
Note that
N \ A = ∪i∈Zk ([i] \ Ai )
is an infinite set, so there must exist some j ∈ Zk such that [j] \ Aj is an infinite
set, i.e., Aj = ∅. This implies that G is a proper subgroup of Zk , so G = dZk for
some d > 1 and d | k.
We conclude that A = ∪i∈G Ai ⊆ {nd : n ∈ N}.

4629. Proposed by Minh Nguyen.

Determine the smallest natural number n such that in any n-element subset of
{1, 2, . . . , 2020} there exist four different a,b,c,d satisfying a + 2b + 3c + 4d ≤ 2020.
We received 9 submissions and 8 of them were complete and correct. We present
the solution by the majority of the solvers.
Using the rearrangement inequality, the minimum value of S := a + 2b + 3c + 4d
is achieved when a > b > c > d. Because the n-element subset could also include
the greatest n numbers from the set, the minimum value of S is achieved when
d is the smallest number in the subset and (a, b, c, d) = (a, a − 1, a − 2, a − 3).
Therefore Smin = 10a − 20 and the inequality S ≤ 2020 leads to a = 204 (and thus
d = 201) being the smallest value of a in the subset of the greatest n numbers
from the set for which the given inequality is satisfied. Thus, the smallest such n
is 2020 − 200 = 1820.

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406/ Solutions

4630. Proposed by George Stoica.

We consider an equilateral triangle ABC with the circumradius 1 and a point D
on or inside its circumcircle. Prove that 3 ≤ AD + BD + CD ≤ 4.
There were 8 correct and 4 incomplete solutions submitted. We present two solu-
tions.
Solution 1, using the ingredients of several submissions.
Let O be the centre of the circumcircle of triangle ABC, P be the midpoint of
BC, and AQ be the diameter of the circle through P . By symmetry, it suffices to
consider points in the sector OBC of the circumcircle.
Observe that R, the sector of the circle bounded by the chord and short arc joining
B and C is reflected with axis BC to a region S that contains triangle OBC. Let
X ∈ R and Y ∈ S be corresponding points with respect to the reflection. Since
∠AY X exceeds 90◦ , AY ≤ AX whence

AY + BY + CY ≤ AX + BX + CX.

Therefore, to find the maximizing position of D, we need only look in R while to

find the minimizing position, in S.
A

W Y
P
B C

Z X
Q

For the maximizing problem, suppose that D is at position X in R. Let CX

produced meet the arc BC at Z. Since ∠AZX = ∠AZC = 60◦ and ∠ZAX ≤ 60◦ ,
∠AXZ ≥ 60◦ and so AZ ≥ AX. Hence

AX + BX + CX ≤ AX + BZ + ZX + CX
≤ AZ + BZ + CZ
= 2AZ ≤ 2AQ = 4,

with equality if and only if X = Z = Q.

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 Solutions /407

The last inequality follows since AQ is a diameter of the circumcircle. The fact
that BZ + CZ = AZ needs more argument. Let W be located on AZ so that
W Z = BZ. Since ∠AZB = ∠ACB = 60◦ , triangle BW Z is equilateral and
W B = BZ. Since ∠BAW = ∠BCZ and

∠BW A = 180◦ − ∠BW Z = 120◦ = ∠BZC,

it follows that ∠ABW = ∠CBZ. Because BA = BC and BW = BZ, triangles

ABW and CBZ are conguent (SAS) and so

BZ + CZ = ZW + W A = AZ.

For the minimizing problem, suppose that D lies within the triangle ABC. A 60◦
rotation with centre B takes C to A, A to E and D to F , so that AEBC is a
60◦ −120◦ rhombus. Since BF = BD, triangle BDF is equilateral and BD = F D.
Also AD = EF . Hence

AD + BD + CD = EF + F D + DC ≥ EC.

But CE right bisects AB, and so CE is equal to twice the altitude of triangle
ABC, namely 3. Equality occurs if D coincides with the circumcenter O.
A
E

D
B C

Solution 2, by UCLan Cyprus Problem Solving Group.

We need consider only points in the sector OBC. Consider first the maximization
problem. If D lies within the triangle OBC, then an argument similar to that in
Solution 1 establishes that

BD + CD ≤ BO + CO = 2,

Since AD is less than a diameter of the circumcircle,

AD + BD + CD ≤ 2 + 2 = 4.

Now let D lie in the segment bounded by the chord and arc BC, and let AD
intersect the circle again at G. By Ptolemy’s theorem

AB · CG + AC · BG = BC · AG whence CG + BG = AG.

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Therefore
AD + BD + CD ≤ AG + BG + CG = 2AG ≤ 4,
with equality if and only if D = G and AG is a diameter.
−−→
For the minimization problem, let x = OX for X = A, B, C, D and corresponding
lower cases. Then

(a − d) · a ≤ ka − dkkak = ka − dk = AD

with similar assertions for b and c. Then

AD + BD + CD ≥ (a − d) · a + (b − d) · b + (c − d) · c
= 3 − d · (a + b + c)
=3−d·0
= 3,

since ABC is an equilateral triangle.

Crux Mathematicorum, Vol. 47(8), October 2021