December 4th
Today I learned that in all nonarchimedean fields $K$ with absolute value $|\bullet|,$ all triangles are isosceles. This was a lemma in a larger result which was kind of disgusting to state in the midst of a homework problem. Anyways, the statement that all triangles are isosceles is really saying that if $|x|\ne|y|,$ then\[|x+y|=\max\{|x|,|y|\}.\]That is, for a triangle with side lengths $x,$ $y,$ and $x+y$ (translate a triangle to one with vertices at the origin, $x,$ and $-y$) we have that two of the sides must always be the same. That is, if $|x|=|y|,$ then we're done; else, the side $|x+y|$ is equal to one of the other two.
Let's prove this quickly. Without loss of generality, $|x| \gt |y|$ so that we want to show $|x+y|=|x|.$ Because our absolute value is nonarchimedean, we already know that\[|x+y|\le\max\{|x|,|y|\}=|x|.\]But in the other direction, we may write\[|x|=|(x+y)+(-y)|\le\max\{|x+y|,|y|\}.\]The maximum looks ambiguous, but we know that $|x| \gt |y|,$ so the statement $|x| \lt |y|$ is impossible, so $|x+y|$ must instead be the maximum. It follows $|x|\le|x+y|,$ from which the equality $|x|=|x+y|$ follows.
This result generalizes the fact that all triangles are isosceles in $\QQ_p$ to any field equipped with a nonarchimedean absolute value. Explicitly, in $\QQ_p,$ the statement that $|x|_p\ne|y|_p$ implies\[|x+y|_p=\max\{|x|_p,|y|_p\}\]is really saying that $\nu_p(x+y)=\min\{\nu_p(x),\nu_p(y)\}$ when $\nu_p(x)\ne\nu_p(y).$ Here, we'll say $\nu_p(x) \lt \nu_p(y)$ without loss of generality. Note if $x=0,$ then done. Writing $x=p^\alpha a$ and $y=p^\beta b$ with $\alpha=\nu_p(x) \gt \beta=\nu_p(y),$ we have that\[p^\alpha a+p^\beta b=p^\alpha\left(a+p^{\beta-\alpha}b\right),\]and $p$ does not divide the second factor because $\beta-\alpha \gt 0.$ So indeed, all triangles are isosceles in $\QQ_p,$ and we can attempt to push the proof a bit harder to generalize to the above, but it's not super conducive.