It looks like the numerical value $\epsilon=1/2024$ matters and that scares me. Also, may you please post the other problems too?

Well, from empirical observations, it might even be true with $\varepsilon \approx 0.45$, but it is very much conceivable that there is a solution which produces a value around $\frac{1}{2024}$ (and that there are other solutions which produce even smaller values). Note that it is not even obvious that this is true for any fixed value of $\varepsilon$.

Disclaimer: The proposers of this problem spent many hours on this problem and concluded that any choice of $\epsilon\le 1-\frac{2}{\sqrt{6}}$ will have similar difficulty. The only reason $\epsilon=\frac{1}{2024}$ was given was to intimidate the students and as an extent of the P2 troll(or not troll).

MNJ2357 wrote: Disclaimer: The proposers of this problem spent many hours on this problem and concluded that any choice of $\epsilon\le 1-\frac{2}{\sqrt{6}}$ will have similar difficulty. The only reason $\epsilon=\frac{1}{2024}$ was given was to intimidate the students and as an extent of the P2 troll(or not troll). can you give an outline of the solution, plz?

bumpity bump

bumpity bump bump

does anyone have a solution

bumpity bump bump bUmP

bumpity bump bump bUmP BuMp

Seems very interesting, anyone?

Here's a proof of the bound given in #4. Note that $\triangle XYZ$ is the first Morley triangle, which is equilateral with sidelengths $8R \sin\left(\frac{A}{3}\right)\sin\left(\frac{B}{3}\right) \sin\left(\frac{C}{3}\right)$. By here we can conclude that the incenter lies inside $\triangle XYZ$. It remains to show that $(1 - \varepsilon) \cdot r \ge 8R \sin\left(\frac{A}{3}\right)\sin\left(\frac{B}{3}\right) \sin\left(\frac{C}{3}\right)$. Substituting the relation for $r = 4R\sin\left(\frac{A}{2}\right)\sin\left(\frac{B}{2}\right) \sin\left(\frac{C}{2}\right) $, we want to show that \[ (1 - \varepsilon) \cdot \sin\left(\frac{A}{2}\right)\sin\left(\frac{B}{2}\right) \sin\left(\frac{C}{2}\right) \ge 2 \sin\left(\frac{A}{3}\right)\sin\left(\frac{B}{3}\right) \sin\left(\frac{C}{3}\right) \] Now consider $K = \prod_{\text{cyc}} \frac{\sin(A/2)}{\sin(A/3)}$. Note that $f(x) = \frac{\sin(x/2)}{\sin(x/3)}$ is strictly decreasing on $(0, \pi)$. If any of $A$ is larger than $\frac{\pi}{2}$, then \[ K \ge f(\pi) \cdot \min_{x + y = \frac{\pi}{2}} f(x)f(y) \ge \frac{2}{\sqrt{3}} \cdot \frac{3}{\sqrt{2}} = \sqrt{6} \]where the value is minimal at $\{x, y\} = \{0, \frac{\pi}{2}\}$ (this is easy to see graphically but there's no immediate nice proof of it). Else, $K \ge f\left(\frac{\pi}{2}\right)^3 = 2\sqrt{2}$. As such, we get that $\varepsilon \le 1 - \frac{2}{K} \le 1 - \frac{2}{\sqrt{6}}$. (The weaker bound $K \ge \frac{4}{\sqrt{3}}$ suffices for the original problem).