% © Evan Chen
% Downloaded from https://web.evanchen.cc/
\documentclass[11pt]{scrartcl}
\usepackage[sexy]{evan}
\ihead{\footnotesize\textbf{\thetitle}}
\ohead{\footnotesize\href{http://web.evanchen.cc}{\ttfamily web.evanchen.cc},
updated \today}
\title{EGMO 2019 Solution Notes}
\date{\today}
\begin{document}
\maketitle
\begin{abstract}
This is a compilation of solutions
for the 2019 EGMO.
The ideas of the solution are a mix of my own work,
the solutions provided by the competition organizers,
and solutions found by the community.
However, all the writing is maintained by me.
These notes will tend to be a bit more advanced and terse than the ``official''
solutions from the organizers.
In particular, if a theorem or technique is not known to beginners
but is still considered ``standard'', then I often prefer to
use this theory anyways, rather than try to work around or conceal it.
For example, in geometry problems I typically use directed angles
without further comment, rather than awkwardly work around configuration issues.
Similarly, sentences like ``let $\mathbb{R}$ denote the set of real numbers''
are typically omitted entirely.
Corrections and comments are welcome!
\end{abstract}
\tableofcontents
\newpage
\addtocounter{section}{-1}
\section{Problems}
\begin{enumerate}[\bfseries 1.]
\item %% Problem 1
Find all triples $(a, b, c)$ of real numbers such that $ab + bc + ca = 1$ and
\[ a^2b + c = b^2c + a = c^2a + b. \]
\item %% Problem 2
Let $n$ be a positive integer.
Dominoes are placed on a $2n \times 2n$ board
in such a way that every cell of the board is
(orthogonally) adjacent to exactly one cell covered by a domino.
For each $n$, determine the largest number of dominoes
that can be placed in this way.
\item %% Problem 3
Let $ABC$ be a triangle such that $\angle CAB > \angle ABC$,
and let $I$ be its incenter.
Let $D$ be the point on segment $BC$
such that $\angle CAD = \angle ABC$.
Let $\omega$ be the circle tangent to $AC$ at $A$
and passing through $I$.
Let $X$ be the second point of intersection of $\omega$
and the circumcircle of $ABC$.
Prove that the angle bisectors of $\angle DAB$ and $\angle CXB$
intersect at a point on line $BC$.
\item %% Problem 4
Let $ABC$ be a triangle with incentre $I$.
The circle through $B$ tangent to $AI$ at $I$
meets side $AB$ again at $P$.
The circle through $C$ tangent to $AI$ at $I$
meets side $AC$ again at $Q$.
Prove that $PQ$ is tangent to the incircle of $ABC$.
\item %% Problem 5
Let $n\ge 2$ be an integer,
and let $a_1, a_2, \dots , a_n$ be positive integers.
Show that there exist positive integers $b_1, b_2, \dots, b_n$
satisfying the following three conditions:
\begin{enumerate}[(a)]
\ii $a_i\le b_i$ for $i=1, 2, \dots, n$;
\ii the remainders of $b_1$, $b_2$, \dots, $b_n$
on division by $n$ are pairwise different,
\ii $b_1 + \dots + b_n \le n \left( \frac{n-1}{2}
+ \left\lfloor \frac{a_1 + \dots + a_n}{n} \right\rfloor \right)$.
\end{enumerate}
\item %% Problem 6
On a circle, Alina draws $2019$ chords, the endpoints of which are all different.
A point is considered marked if it is either
\begin{enumerate}[(i)]
\ii one of the $4038$ endpoints of a chord; or
\ii an intersection point of at least two chords.
\end{enumerate}
Of the $4038$ points meeting criterion (i),
Alina labels $2019$ points with a $0$ and the other $2019$ points with a $1$.
She labels each point meeting criterion (ii)
with an arbitrary integer (not necessarily positive).
Along each chord, Alina considers the segments connecting two consecutive marked points.
(A chord with $k$ marked points has $k-1$ such segments.)
She labels each such segment in yellow with the sum of the labels of its two endpoints
and in blue with the absolute value of their difference.
Alina finds that the $N + 1$ yellow labels take each value $0, 1, \dots, N$ exactly once.
Show that at least one blue label is a multiple of $3$.
\end{enumerate}
\pagebreak
\section{Solutions to Day 1}
\subsection{EGMO 2019/1, proposed by Netherlands}
\textsl{Available online at \url{https://aops.com/community/p12141493}.}
\begin{mdframed}[style=mdpurplebox,frametitle={Problem statement}]
Find all triples $(a, b, c)$ of real numbers such that $ab + bc + ca = 1$ and
\[ a^2b + c = b^2c + a = c^2a + b. \]
\end{mdframed}
Answer: $(\pm 1/\sqrt3, \pm 1/\sqrt3, \pm1/\sqrt3)$
where the signs correspond,
and $(\pm1, \pm1, 0)$ and permutations where the signs correspond.
These work and we prove that is all.
We begin by eliminating the condition
via homogenization:
the first equality now reads
\begin{align*}
a^2b+c\left[ ab+bc+ca \right] &= b^2c+a\left[ ab+bc+ca \right] \\
\iff c^2(b+a) &= c(b^2+a^2) \\
\iff c&=0 \text{ or } a^2+b^2 = c(a+b).
\end{align*}
Cyclic variations hold.
So we have two cases.
\begin{itemize}
\ii If any of the variables is zero,
say $a = 0$, then the other two are nonzero.
So from $b^2 = bc$ we get $b = c$ giving
$(\pm1, \pm1, 0)$.
\ii Now assume all three variables are nonzero,
so $a^2+b^2 = c(a+b)$.
If we sum cyclically we get
\[ 2(a^2+b^2+c^2) = 2(ab+bc+ca)
\iff (a-b)^2 + (b-c)^2 + (c-a)^2 = 0 \]
which forces $a=b=c$ and gives the last solution.
\end{itemize}
\pagebreak
\subsection{EGMO 2019/2, proposed by Luxembourg}
\textsl{Available online at \url{https://aops.com/community/p12141498}.}
\begin{mdframed}[style=mdpurplebox,frametitle={Problem statement}]
Let $n$ be a positive integer.
Dominoes are placed on a $2n \times 2n$ board
in such a way that every cell of the board is
(orthogonally) adjacent to exactly one cell covered by a domino.
For each $n$, determine the largest number of dominoes
that can be placed in this way.
\end{mdframed}
The answer is $\binom{n+1}{2}$ and a construction is shown below.
For each domino, its \emph{aura} consists of all the cells
which are adjacent to a cell of the domino.
There are up to eight squares in each aura,
but some auras could be cut off by the boundary of the board,
which means that there could be as few as five squares.
So we want to estimate how many auras get cut off.
We denote by $a$, $b$, $c$, $k$ the number of auras
with $5$, $6$, $7$, $8$ cells on the boundary,
so we want an upper bound on $a+b+c+k$.
Note also that $\boxed{a \le 4}$
since such a cell uses a corner of the grid.
\begin{center}
\begin{asy}
size(11cm);
path aura = (1,0)--(2,0)--(2,1)--(3,1)--(3,3)--(2,3)--(2,4)--(1,4)--(1,3)--(0,3)--(0,1)--(1,1)--cycle;
picture aura_up(pen fill) {
picture pic = new picture;
fill(pic, aura, fill);
draw(pic, (1,1)--(2,1), lightgrey );
draw(pic, (1,3)--(2,3), lightgrey );
draw(pic, (0,2)--(3,2), lightgrey );
draw(pic, (1,1)--(1,3), lightgrey );
draw(pic, (2,1)--(2,3), lightgrey );
draw(pic, aura, blue+1.25);
return pic;
}
picture aura_right(pen fill) {
return shift(0,3) * rotate(-90) * aura_up(fill);
}
add(shift(-1,-1)*aura_up(palegreen));
add(shift(-1,3)*aura_up(palegreen));
add(shift(-1,7)*aura_up(palegreen));
add(shift(1,1)*aura_up(palegreen));
add(shift(1,5)*aura_up(palegreen));
add(shift(3,3)*aura_up(palegreen));
add(shift(6,3)*aura_up(palegreen));
add(shift(8,1)*aura_up(palegreen));
add(shift(8,5)*aura_up(palegreen));
add(shift(10,-1)*aura_up(palegreen));
add(shift(10,3)*aura_up(palegreen));
add(shift(10,7)*aura_up(palegreen));
add(shift(4,6)*aura_right(palered));
add(shift(2,8)*aura_right(palered));
add(shift(6,8)*aura_right(palered));
/*
add(shift(0,10)*aura_right(palered));
add(shift(4,10)*aura_right(palered));
add(shift(8,10)*aura_right(palered));
*/
add(shift(4,1)*aura_right(palered));
add(shift(2,-1)*aura_right(palered));
add(shift(6,-1)*aura_right(palered));
add(shift(0,-3)*aura_right(palered));
add(shift(4,-3)*aura_right(palered));
add(shift(8,-3)*aura_right(palered));
draw(shift(0,-2)*scale(12)*unitsquare, black+2);
path cell = shift(16,-2)*unitsquare;
for (int i = 0; i <= 11; ++i) {
for (int j = 0; j <= 11; ++j) {
if (max(abs(i-5.5), abs(j-5.5)) == 5.5)
filldraw(shift(i,j)*cell, paleblue, grey);
/*
else if (max(abs(i-5.5), abs(j-5.5)) == 3.5)
filldraw(shift(i,j)*cell, paleblue, grey);
else if (max(abs(i-5.5), abs(j-5.5)) == 1.5)
filldraw(shift(i,j)*cell, paleblue, grey);
*/
else draw(shift(i,j)*cell, grey);
}
}
\end{asy}
\end{center}
The big observation about the auras on the edge:
\begin{claim*}
The auras of type $a$, $b$, $c$ have
$4$, $4$, and $3$-$4$ cells on the boundary of the grid, respectively.
(The boundary is the $4(2n-1)$ cells touching an edge of the board.)
\end{claim*}
Consequently, we have a bound
$4a + 4b + 3c \le 4(2n-1)$.
On the other hand, we obviously have
$5a + 6b + 7c + 8k = 4n^2$.
Therefore,
\begin{align*}
4n^2 + 2(2n-1) &\ge (5a+6b+7c+8k) + (2a+2b+1.5c) \\
&= 8(a+b+c+k) + 0.5c - a \\
&\ge 8(a+b+c+k) + 0 - 4 \\
\implies \frac{n(n+1)}{2} + \frac 14 &\ge a+b+c+k
\end{align*}
which implies the desired bound after taking the floor of left-hand side.
\begin{remark*}
In fact IMO 1999/3 shows the reverse bound:
we now give a proof that every tiling in this problem
has \emph{exactly} $\half n (n+1)$ dominoes.
Color the board as shown below into ``rings''.
\begin{center}
\begin{asy}
size(5cm);
path cell = unitsquare;
for (int i = 0; i <= 11; ++i) {
for (int j = 0; j <= 11; ++j) {
if (max(abs(i-5.5), abs(j-5.5)) == 5.5)
filldraw(shift(i,j)*cell, paleblue, grey);
else if (max(abs(i-5.5), abs(j-5.5)) == 3.5)
filldraw(shift(i,j)*cell, paleblue, grey);
else if (max(abs(i-5.5), abs(j-5.5)) == 1.5)
filldraw(shift(i,j)*cell, paleblue, grey);
else draw(shift(i,j)*cell, grey);
}
}
\end{asy}
\end{center}
Every aura covers exactly four blue cells. Done.
\end{remark*}
\pagebreak
\subsection{EGMO 2019/3, proposed by Poland}
\textsl{Available online at \url{https://aops.com/community/p12141505}.}
\begin{mdframed}[style=mdpurplebox,frametitle={Problem statement}]
Let $ABC$ be a triangle such that $\angle CAB > \angle ABC$,
and let $I$ be its incenter.
Let $D$ be the point on segment $BC$
such that $\angle CAD = \angle ABC$.
Let $\omega$ be the circle tangent to $AC$ at $A$
and passing through $I$.
Let $X$ be the second point of intersection of $\omega$
and the circumcircle of $ABC$.
Prove that the angle bisectors of $\angle DAB$ and $\angle CXB$
intersect at a point on line $BC$.
\end{mdframed}
Here is a cross-ratio/trig solution:
rare for me to do one of these,
but this problem called out to me that way.
As usual, let $\alpha = \angle BAC$, etc.
Let $T$ be the foot of the bisector of $\angle BAD$ on $\ol{BD}$,
so that
\[ \frac{TB}{TC} = \frac{AB \sin \angle BAT}{AC \sin \angle CAT}
= \frac{AB \sin \frac{\alpha-\beta}{2}}
{AC \sin \frac{\alpha+\beta}{2}}. \]
Also, call $(ABC)$ by $\Gamma$ and let ray $XI$ meet $\Gamma$ again at $W$,
meaning that $\angle WXA = \angle IXA = \angle IAC = \frac{\beta}{2}$,
since we assume $\ol{AC}$ was tangent to $(IAX)$.
Thus arc $\widehat{AW}$ also has measure $\beta$.
\begin{center}
\begin{asy}
pair A = dir(50);
pair B = dir(210);
pair C = dir(330);
pair I = incenter(A, B, C);
pair M_a = circumcenter(B, I, C);
pair M_b = circumcenter(C, I, A);
pair M_c = circumcenter(A, I, B);
pair W = A*B/M_a;
pair X = -W+2*foot(origin, I, W);
pair D = extension(B, C, A, C*C/A);
filldraw(unitcircle, opacity(0.1)+lightcyan, lightblue);
draw(A--B--C--cycle, lightblue);
draw(X--W, lightred);
draw(B--M_b, lightred);
draw(C--M_c, lightred);
draw(A--M_a, lightred);
filldraw(W--M_b--A--M_c--cycle, opacity(0.1)+yellow, deepgreen);
filldraw(X--B--M_a--C--cycle, opacity(0.1)+yellow, deepgreen);
draw(A--D, lightblue);
pair T = extension(A, incenter(A, B, D), B, D);
draw(X--T--A, dashed+brown);
dot("$A$", A, dir(A));
dot("$B$", B, dir(B));
dot("$C$", C, dir(C));
dot("$I$", I, dir(I));
dot("$M_a$", M_a, dir(M_a));
dot("$M_b$", M_b, dir(M_b));
dot("$M_c$", M_c, dir(M_c));
dot("$W$", W, dir(W));
dot("$X$", X, dir(X));
dot("$D$", D, dir(D));
dot("$T$", T, dir(T));
/* TSQ Source:
A = dir 50
B = dir 210
C = dir 330
I = incenter A B C
M_a = circumcenter B I C
M_b = circumcenter C I A
M_c = circumcenter A I B
W = A*B/M_a
X = -W+2*foot origin I W
D = extension B C A C*C/A
unitcircle 0.1 lightcyan / lightblue
A--B--C--cycle lightblue
X--W lightred
B--M_b lightred
C--M_c lightred
A--M_a lightred
W--M_b--A--M_c--cycle 0.1 yellow / deepgreen
X--B--M_a--C--cycle 0.1 yellow / deepgreen
A--D lightblue
T = extension A incenter A B D B D
X--T--A dashed brown
*/
\end{asy}
\end{center}
Now,
\begin{align*}
\frac{XB}{XC}
&= -(BC;XM_a)_\Gamma
\overset{I}{=} -(M_b M_c; W A)_\Gamma \\
&= -\frac%
{\sin \half \mathrm{m}\widehat{M_b W}}
{\sin \half \mathrm{m}\widehat{M_c W}}
\div \frac%
{\sin \half \mathrm{m}\widehat{M_b A}}
{\sin \half \mathrm{m}\widehat{M_c A}}
= \frac{\sin\frac{\alpha-\beta}{2}}
{\sin\frac{\alpha +\gamma}{2}}
\div \frac%
{\sin \frac{\gamma}{2}}
{\sin \frac{\beta}{2}} \\
\end{align*}
Therefore,
\[
\frac{XB}{XC} \div \frac{TB}{TC}
= \frac{\sin\frac{\alpha+\beta}{2} \sin\frac{\gamma}{2}}%
{\sin\frac{\alpha+\gamma}{2} \sin\frac{\beta}{2}}
\div \frac{AB}{AC} \\
= \frac{2\cos \frac{\gamma}{2} \sin \frac{\gamma}{2}}%
{2\cos \frac{\beta}{2} \sin \frac{\beta}{2}} \div \frac{AB}{AC}
= 1
\]
the last step being the law of sines $AB/AC = \sin\gamma/\sin\beta$.
\pagebreak
\section{Solutions to Day 2}
\subsection{EGMO 2019/4, proposed by Poland}
\textsl{Available online at \url{https://aops.com/community/p12146839}.}
\begin{mdframed}[style=mdpurplebox,frametitle={Problem statement}]
Let $ABC$ be a triangle with incentre $I$.
The circle through $B$ tangent to $AI$ at $I$
meets side $AB$ again at $P$.
The circle through $C$ tangent to $AI$ at $I$
meets side $AC$ again at $Q$.
Prove that $PQ$ is tangent to the incircle of $ABC$.
\end{mdframed}
Let $E$ and $F$ be the tangency points of the incircle on $AC$ and $AB$.
By angle chasing,
\[ \angle PIF = \angle AIF - \angle AIP
= \left( 90\dg - \half \angle A \right) - \half \angle B = \half \angle C. \]
Similarly, $\angle EIQ = \half \angle B$.
\begin{center}
\begin{asy}
pair A = dir(110);
pair B = dir(210);
pair C = dir(330);
pair I = incenter(A, B, C);
pair D = foot(I, B, C);
pair E = foot(I, C, A);
pair F = foot(I, A, B);
pair _P = abs(A-I)*abs(A-I)/abs(A-B) * dir(B-A) + A;
pair _Q = abs(A-I)*abs(A-I)/abs(A-C) * dir(C-A) + A;
pair P = _P;
pair Q = _Q;
filldraw(incircle(A, B, C), opacity(0.1)+blue, blue);
filldraw(A--B--C--cycle, opacity(0.1)+blue, blue);
draw(P--Q, red);
draw(P--I--F, deepgreen);
draw(E--I--Q, deepgreen);
pair T = foot(I, P, Q);
draw(I--T, red+dashed);
draw(A--I, lightblue);
draw(B--I--C, lightblue);
dot("$A$", A, dir(A));
dot("$B$", B, dir(B));
dot("$C$", C, dir(C));
dot("$I$", I, dir(270));
dot("$D$", D, dir(D));
dot("$E$", E, dir(E));
dot("$F$", F, dir(F));
dot("$P$", P, dir(P));
dot("$Q$", Q, dir(Q));
dot("$T$", T, dir(T));
/* -----------------------------------------------------------------+
| TSQX: by CJ Quines and Evan Chen |
| https://github.com/vEnhance/dotfiles/blob/main/py-scripts/tsqx.py |
+-------------------------------------------------------------------+
A = dir 110
B = dir 210
C = dir 330
I 270 = incenter A B C
D = foot I B C
E = foot I C A
F = foot I A B
!pair _P = abs(A-I)*abs(A-I)/abs(A-B) * dir(B-A) + A;
!pair _Q = abs(A-I)*abs(A-I)/abs(A-C) * dir(C-A) + A;
P = _P
Q = _Q
incircle A B C / 0.1 blue / blue
A--B--C--cycle / 0.1 blue / blue
P--Q / red
P--I--F / deepgreen
E--I--Q / deepgreen
T = foot I P Q
I--T / red dashed
A--I / lightblue
B--I--C / lightblue
*/
\end{asy}
\end{center}
Let $T_p$, $T_q$ denote the second tangency of $P$, $Q$ to the incircle.
Then $\angle E I T_q = \angle B$ and $\angle T_p I F = \angle C$.
Since $\angle E I F = \angle B + \angle C$, it follows $T_p = T_q$.
\begin{remark*}
Notice we really only need $\angle PIQ = 90\dg - \half \angle A$
by essentially the same argument.
\end{remark*}
\pagebreak
\subsection{EGMO 2019/5, proposed by Merlijn Staps (NLD)}
\textsl{Available online at \url{https://aops.com/community/p12146863}.}
\begin{mdframed}[style=mdpurplebox,frametitle={Problem statement}]
Let $n\ge 2$ be an integer,
and let $a_1, a_2, \dots , a_n$ be positive integers.
Show that there exist positive integers $b_1, b_2, \dots, b_n$
satisfying the following three conditions:
\begin{enumerate}[(a)]
\ii $a_i\le b_i$ for $i=1, 2, \dots, n$;
\ii the remainders of $b_1$, $b_2$, \dots, $b_n$
on division by $n$ are pairwise different,
\ii $b_1 + \dots + b_n \le n \left( \frac{n-1}{2}
+ \left\lfloor \frac{a_1 + \dots + a_n}{n} \right\rfloor \right)$.
\end{enumerate}
\end{mdframed}
Note that if $a_i > n$,
we can replace $a_i$ with $a_i-n$ and $b_i$ with $b_i-n$,
and nothing changes.
So we may as well assume $a_i \in \{1, \dots, n\}$ for each $i$.
We choose our $b_i$'s in the following way.
Draw an $n \times n$ grid and in the $i$th
column fill in the bottom $a_i-1$ cells red.
We can select $b_i$ by marking $n$ cells,
one in each row or column.
If we chose $j$th lowest row in the $i$th column,
then we would set $b_i = j$ on non-red cells
and $b_i = j + n$ on red cells.
In this way, define the \emph{penalty} $p$ as
the number of selected cells which are red.
Then
\[ b_1 + \dots + b_n = (1+2+\dots+n) + n \cdot p
= n \cdot \frac{n-1}{2} + n \cdot (p+1). \]
and we seek to minimize the penalty $p$.
\begin{center}
\begin{asy}
size(8cm);
fill( (1,0)--(1,2)--(2,2)--(2,3)--(4,3)--(4,4)--(5,4)--(5,0)--cycle, palered);
for (int i=0; i<=5; ++i) {
draw((i,0)--(i,5), grey);
draw((0,i)--(5,i), grey);
}
draw(scale(5)*unitsquare);
label("$b_i \equiv 1 \pmod 5$", (0,0.5), dir(180), lightblue);
label("$b_i \equiv 2 \pmod 5$", (0,1.5), dir(180), lightblue);
label("$b_i \equiv 3 \pmod 5$", (0,2.5), dir(180), lightblue);
label("$b_i \equiv 4 \pmod 5$", (0,3.5), dir(180), lightblue);
label("$b_i \equiv 5 \pmod 5$", (0,4.5), dir(180), lightblue);
label("$a_1-1$", (0.5,0), dir(-90), lightred);
label("$a_2-1$", (1.5,0), dir(-90), lightred);
label("$a_3-1$", (2.5,0), dir(-90), lightred);
label("$a_4-1$", (3.5,0), dir(-90), lightred);
label("$a_5-1$", (4.5,0), dir(-90), lightred);
label("$b_1$", (0.5, 3.5), blue);
label("$b_2$", (1.5, 2.5), blue);
label("$b_3$", (2.5, 4.5), blue);
label("$b_4$", (3.5, 0.5), blue);
label("$b_5$", (4.5, 1.5), blue);
\end{asy}
\end{center}
But the expected penalty of a \emph{random} permutation
is the red area divided by $n$, which is \[ \mathbb E[p] = \frac{(a_1-1) + \dots + (a_n-1)}{n} \]
and so there exists a choice for which
the penalty is at most $\left\lfloor \mathbb E[p] \right\rfloor$.
This gives the required result.
\begin{remark*}
The visual aid can be excised from the solution;
which can then be rewritten more tersely as follows.
After assuming $1 \le a_i \le n$ for each $n$,
pick a uniformly random permutation $\sigma$ on
$\{1, \dots, n\}$ and define
\[ b_i =
\begin{cases}
n + \sigma(i) & a_i > \sigma(i) \\
\sigma(i) & \text{otherwise}.
\end{cases}
\]
As before $\mathbb E[e_\sigma] = \sum_{i=1}^n \frac{a_i-1}{n}$
and the rest is the same.
\end{remark*}
\pagebreak
\subsection{EGMO 2019/6, proposed by United Kingdom}
\textsl{Available online at \url{https://aops.com/community/p12146850}.}
\begin{mdframed}[style=mdpurplebox,frametitle={Problem statement}]
On a circle, Alina draws $2019$ chords, the endpoints of which are all different.
A point is considered marked if it is either
\begin{enumerate}[(i)]
\ii one of the $4038$ endpoints of a chord; or
\ii an intersection point of at least two chords.
\end{enumerate}
Of the $4038$ points meeting criterion (i),
Alina labels $2019$ points with a $0$ and the other $2019$ points with a $1$.
She labels each point meeting criterion (ii)
with an arbitrary integer (not necessarily positive).
Along each chord, Alina considers the segments connecting two consecutive marked points.
(A chord with $k$ marked points has $k-1$ such segments.)
She labels each such segment in yellow with the sum of the labels of its two endpoints
and in blue with the absolute value of their difference.
Alina finds that the $N + 1$ yellow labels take each value $0, 1, \dots, N$ exactly once.
Show that at least one blue label is a multiple of $3$.
\end{mdframed}
Only the labels mod $3$ matter at all.
Assume for contradiction no blue labels are divisible by $3$.
Let $e_{ij}$ denote the number of segments
joining $i \pmod 3$ to $j \pmod 3$.
By double-counting (noting that points in (ii) are
counted an even number of times
but points in (i) are counted once)
we derive that
\begin{align*}
e_{01} + e_{02} &\equiv 2019 \pmod 2 \\
e_{01} + e_{12} &\equiv 2019 \pmod 2 \\
e_{02} + e_{12} &\equiv 0 \pmod 2
\end{align*}
which gives \[ e_{02} \equiv e_{12} \equiv 1 - e_{01} \pmod 2. \]
However, one can check this is incompatible
with the hypothesis that the yellow labels
are $0$, $1$, \dots, $N$.
\begin{remark*}
In addition, we can replace the points/segments
by any graph $G$ for which there are
\begin{itemize}
\ii an odd number of leaves (or just odd-degree vertices) labeled $0$,
\ii an odd number of leaves (or just odd-degree vertices) labeled $1$,
\ii and the remaining vertices have even degree.
\end{itemize}
Thus the geometry of the problem is smoke and mirrors, too.
\end{remark*}
\pagebreak
\end{document}