Problem: Let $ABC$ be a triangle such that
\[
  \left( \cot \frac{A}{2} \right)^2
  + \left( 2\cot \frac{B}{2} \right)^2
  + \left( 3\cot \frac{C}{2} \right)^2
  = \left( \frac{6s}{7r} \right)^2,
\]
where $s$ and $r$ denote its semiperimeter and its inradius, respectively.
Prove that triangle $ABC$ is similar to a triangle $T$ whose side lengths are all positive integers
with no common divisors and determine these integers.

Solution: Let $x = s-a$, $y = s-b$, $z = s-c$ in the usual fashion, then the equation reads
\[ x^2 + 4y^2 + 9z^2 = \left( \frac67(x+y+z) \right)^2. \]
However, by Cauchy-Schwarz, we have
\[ \left( 1 + \tfrac14 + \tfrac19 \right)\left( x^2 + 4y^2 + 9z^2 \right) \ge \left( x+y+z \right)^2  \]
with equality if and only if $1 : \tfrac12 : \tfrac13 = x : 2y : 3z$,
id est $x : y : z = 1 : \tfrac14 : \tfrac19 = 36 : 9 : 4$.
This is equivalent to $y+z : z+x : x+y = 13 : 40 : 45$.

%Wow.
%I was laughing at how un-geometric this problem looked,
%as it was the only geometry problem on the USAMO 2002.
%Then I actually tried it and realized it was not geometry.

\begin{remark*}
You can tell this is not a geometry problem
because you eliminate the cotangents right away to get an algebra problem\dots
and then you realize the problem claims that one equation
can determine three variables up to scaling,
at which point you realize it has to be an inequality
(otherwise degrees of freedom don't work).
So of course, Cauchy-Schwarz\dots
\end{remark*}