Marinada '25

[lang=hr] Ajme što je bio ovaj prijašnji zadatak??. Ajde još nisam pao u nesvijest, idemo vidjeti može li me sljedeći zadatak nokautirati. Možda i može... Neka su $\alpha,\beta,\gamma $ kutovi netupokutnog trokuta, a $R$ i $r$ polumjeri njemu opisane i upisane kružnice. Nađi najveći $x$ tako da uvijek vrijedi nejednakost $$\sec\alpha+\sec\beta+\sec\gamma\geq\frac{xR}{R+r}$$. U polje za odgovore upiši $x^2$. Ako je $x^2$ razlomak zapiši ga u skraćenom obliku $a/b$ te u polje upiši umnožak $a\cdot b$. Ako je rezultat koji bi upisao u polje jako velik izračunaj njegov ostatak pri dijeljenju s $1000007$. [/lang] [lang=en] What was that previous problem? Come on, I haven’t fainted yet; let’s see if the next problem can knock me out. Maybe it can… Let $\alpha,\beta,\gamma$ be the angles of an acute‑angled triangle, and let $R$ and $r$ be the radii of its circumcircle and incircle, respectively. Find the largest constant $x$ such that the inequality $$\sec\alpha+\sec\beta+\sec\gamma \ge \frac{xR}{R+r}$$ holds for every acute‑angled triangle. Enter $x^{2}$ in the answer box. If $x^{2}$ is a fraction, write it in lowest terms as $\displaystyle\frac{a}{b}$ and then enter the product $a\cdot b$ in the answer box. If the resulting number is very large, compute its remainder modulo $1000007$. [/lang]