\documentclass[11pt]{scrartcl}
\usepackage{evan}
\begin{document}
\title{TSTST 2020/3}
\subtitle{Evan Chen}
\author{Twitch Solves ISL}
\date{Episode 39}
\maketitle
\section*{Problem}
We say a nondegenerate triangle whose angles have
measures $\theta_1$, $\theta_2$, $\theta_3$ is \emph{quirky}
if there exists integers $r_1$, $r_2$, $r_3$, not all zero,
such that \[ r_1 \theta_1 + r_2 \theta_2 + r_3 \theta_3 = 0. \]
Find all integers $n \ge 3$ for which a triangle with
side lengths $n-1$, $n$, $n+1$ is quirky.
\section*{Video}
\href{https://www.youtube.com/watch?v=zkygeEZ_scc}{\texttt{https://youtu.be/zkygeEZ\_scc}}
\section*{External Link}
\url{https://aops.com/community/p18933954}
\newpage
\section*{Solution}
The answer is $n = 3,\ 4,\ 5,\ 7$.
We first introduce a variant of the $k$th Chebyshev polynomials
in the following lemma
(which is standard, and easily shown by induction).
\begin{lemma*}
For each $k \ge 0$ there exists $P_k(X) \in \ZZ[X]$,
monic for $k \ge 1$ and with degree $k$, such that
\[ P_k( X + X^{-1} ) \equiv X^k + X^{-k}. \]
The first few are $P_0(X) \equiv 2$,
$P_1(X) \equiv X$,
$P_2(X) \equiv X^2-2$,
$P_3(X) \equiv X^3-3X$.
% In particular, $P_k(2\cos\theta) = 2\cos(k\theta)$.
\end{lemma*}
Suppose the angles of the triangle are $\alpha < \beta < \gamma$,
so the law of cosines implies that
\[ 2\cos\alpha =\frac{n+4}{n+1} \qquad \text{and} \qquad 2\cos\gamma = \frac{n-4}{n-1}. \]
%\begin{align*}
% 2\cos\alpha &= \tfrac{n^2+(n+1)^2-(n-1)^2}{n(n+1)}
% = \frac{n+4}{n+1} \\
% 2\cos\beta &= \tfrac{(n-1)^2+(n+1)^2-n^2}{(n-1)(n+1)}
% = \frac{n^2+2}{n^2-1} \\
% 2\cos\gamma &= \tfrac{n^2+(n-1)^2-(n+1)^2}{n(n-1)}
% = \frac{n-4}{n-1}.
%\end{align*}
\begin{claim*}
The triangle is quirky iff there exists $r,s \in \ZZ_{\ge0}$
not both zero such that
\[ \cos(r\alpha) = \pm \cos(s\gamma)
\qquad \text{or equivalently} \qquad
P_r\left( \frac{n+4}{n+1} \right)
= \pm P_s\left( \frac{n-4}{n-1} \right). \]
\end{claim*}
\begin{proof}
If there are integers $x$, $y$, $z$ for which
$x\alpha + y\beta + z\gamma = 0$,
then we have that $(x-y)\alpha = (y-z)\gamma - \pi y$,
whence it follows that we may take $r = |x-y|$ and $s = |y-z|$
(noting $r=s=0$ implies the absurd $x=y=z$).
Conversely, given such $r$ and $s$
with $\cos(r\alpha) = \pm \cos(s\gamma)$,
then it follows that $r \alpha \pm s \gamma = k\pi
= k(\alpha + \beta + \gamma)$ for some $k$,
so the triangle is quirky.
\end{proof}
If $r=0$, then by rational root theorem on $P_s(X) \pm 2$ it follows
$\frac{n-4}{n-1}$ must be an integer
which occurs only when $n = 4$ (recall $n \ge 3$).
Similarly we may discard the case $s = 0$.
Thus in what follows assume $n \neq 4$ and $r,s > 0$.
Then, from the fact that $P_r$ and $P_s$ are nonconstant
monic polynomials, we find
\begin{corollary*}
If $n \neq 4$ works, then when $\frac{n+4}{n+1}$
and $\frac{n-4}{n-1}$ are written as fractions in lowest terms,
the denominators have the same set of prime factors.
\end{corollary*}
But $\gcd(n+1, n-1)$ divides $2$,
and $\gcd(n+4, n+1)$, $\gcd(n-4,n-1)$ divide $3$.
So we only have three possibilities:
\begin{itemize}
\ii $n+1 = 2^u$ and $n-1 = 2^v$ for some $u,v \ge 0$.
This is only possible if $n = 3$.
Here $2\cos\alpha = \frac74$ and $2\cos\gamma = -\frac12$,
and indeed $P_2(-1/2) = -7/4$.
\ii $n+1 = 3 \cdot 2^u$ and $n-1 = 2^v$ for some $u,v \ge 0$,
which implies $n = 5$.
Here $2\cos\alpha = \frac32$ and $2\cos\gamma = \frac14$,
and indeed $P_2(3/2) = 1/4$.
\ii $n+1 = 2^u$ and $n-1 = 3 \cdot 2^v$ for some $u,v \ge 0$,
which implies $n = 7$.
Here $2\cos\alpha = \frac{11}{8}$ and $2\cos\gamma = \frac12$,
and indeed $P_3(1/2) = -11/8$.
\end{itemize}
Finally, $n=4$ works because the triangle is right,
completing the solution.
\begin{remark*}
[Major generalization due to Luke Robitaille]
In fact one may find all quirky triangles
whose sides are integers in arithmetic progression.
Indeed, if the side lengths of the triangle are $x-y$, $x$, $x+y$
with $\gcd(x,y) = 1$ then the problem becomes
\[ P_r\left( \frac{x+4y}{x+y} \right)
= \pm P_s \left( \frac{x-4y}{x-y} \right) \]
and so in the same way as before,
we ought to have $x+y$ and $x-y$ are both of the form $3 \cdot 2^{\ast}$
unless $rs = 0$.
This time, when $rs=0$, we get the extra solutions $(1,0)$ and $(5,2)$.
For $rs \neq 0$,
by triangle inequality, we have $x-y \le x+y < 3(x-y)$,
and $\min(\nu_2(x-y), \nu_2(x+y)) \le 1$,
so it follows one of $x-y$ or $x+y$ must be in $\{1,2,3,6\}$.
An exhaustive check then leads to
\[ (x,y) \in \left\{ (3,1), (5,1), (7,1), (11,5) \right\}
\cup \left\{ (1,0), (5,2), (4,1) \right\} \]
as the solution set. And in fact they all work.
In conclusion the equilateral triangle,
$3-5-7$ triangle (which has a $120\dg$ angle)
and $6-11-16$ triangle (which satisfies $B=3A+4C$)
are exactly the new quirky triangles (up to similarity)
whose sides are integers in arithmetic progression.
\end{remark*}
\end{document}