After a few internet searches for really hard IMO #6 problems, I found the 2011 IMO #6 problem, which states:
"Let $ABC$ be an acute triangle with circumcircle $\Gamma$. Let $\ell$ be a tangent line to $\Gamma$, and let $\ell_a, \ell_b$ and $\ell_c$ be the lines obtained by reflecting $\ell$ in the lines $BC$, $CA$ and $AB$, respectively. Show that the circumcircle of the triangle determined by the lines $\ell_a, \ell_b$ and $\ell_c$ is tangent to the circle $\Gamma$."
I didn't set out to solve it, but I was intrigued by it, so I opened up Geogebra and started constructing it. Fairly easy constructions led me to see a visual of it, which was great. But then I wondered, "why is one of the assumptions the fact that the starting triangle is acute?" So I moved my initial triangle points around so that the starting triangle is obtuse, acute, right, scalene, isosceles, equilateral...basically everything. The only thing I noticed was that an initial right triangle led to the second circumcircle to be degenerate (just a straight line, although it was still tangent to the first circumcircle). All other triangles led to the second circumcircle being tangent. So...why is that in the initial assumptions? Did I miss something (perhaps a special case)?
Also, I was asked by some students of mine, "who would think of this?" Did the problem's proposer have an idea that this construction would lead to a tangent circle? Was there a deeper question they were investigating that led them to these constructions?