\documentclass[11pt]{scrartcl}
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\begin{document}
\title{Iberoamerican 2021/4}
\subtitle{Evan Chen}
\author{Twitch Solves ISL}
\date{Episode 87}
\maketitle

\section*{Problem}
Let $a$, $b$, $c$, $x$, $y$, $z$ be real numbers such that
\begin{align*}
  a^2+x^2 &= b^2+y^2 = c^2+z^2 = (a+b)^2+(x+y)^2 \\
  &= (b+c)^2+(y+z)^2=(c+a)^2+(z+x)^2
\end{align*}
Show that $a^2+b^2+c^2=x^2+y^2+z^2$.

\section*{Video}
\href{https://www.youtube.com/watch?v=KDnDe6KaqL0&list=PLi6h8GM1FA6yHh4gDk_ZYezmncU1EJUmZ}{\texttt{https://youtu.be/KDnDe6KaqL0}}

\section*{External Link}
\url{https://aops.com/community/p23439842}

\newpage

\section*{Solution}
We present two solutions.

\paragraph{Low IQ solution.}
If the common sum is zero, then all numbers are zero.
So WLOG the common quantity is $1$ by scaling.
Write
\begin{align*}
  1 &= (a+b)^2 + \left( \pm \sqrt{1-a^2} \pm \sqrt{1-b^2} \right)^2 \\
  &= a^2+2ab+b^2 + 2-(a^2+b^2) \pm 2\sqrt{(1-a^2)(1-b^2)} \\
  \implies 2ab+1 &= 2\sqrt{(1-a^2)(1-b^2)} \\
  \implies 4a^2b^2 + 4ab + 1 &= 4(1-a^2)(1-b^2) = 4 - 4a^2 - 4b^2 + 4(ab)^2 \\
  \implies 4ab &= 4(1-a^2)(1-b^2) = 3 - 4a^2 - 4b^2 \\
  \implies a^2+ab+b^2 &= \frac 34.
\end{align*}
Proceeding in the same way we get
\[ b^2+bc+c^2 = \frac34, \quad c^2+ca+a^2 = \frac34. \]
By using HMMT 2014 Problem A-9,
this implies $ab+bc+ca$ is equal to some absolute constant $k$
(not depending on the choice of solution to the system).
Hence, adding gives
\[ a^2+b^2+c^2 = \frac 94 - \half k. \]

Similarly, $x^2+y^2+z^2 = \frac 94 - \half k$, as needed.

\begin{remark*}
  This solution only uses the real hypothesis in the first step
  when scaling the common quantity to $1$.
  Thus, as long as $a^2+x^2 \neq 0$, the problem still holds
  even in the complex setting.

  On the other hand, the tuple $(1,1,1,i,i,i)$
  is a counterexample to the statement without the real hypothesis.
\end{remark*}

\paragraph{High IQ solution.}
As before scale so that the common sum is $1$.
Consider
\[ z_1 = a+xi, \qquad z_2 = b+yi, \qquad z_3 = c+zi \]
in the complex plane.
The problem hypothesis says that the vectors
are on the unit circle such that $|z_i+z_j|=1$ for $i \neq j$,
which implies they are the vertices of an equilateral triangle.
Hence
\[ 0 = z_1^2 + z_2^2 + z_3^2 \]
and the real part of the right-hand side is
$(a^2+b^2+c^2)-(x^2+y^2+z^2)$.

\end{document}
