Question 82
Prove that in every triangle, there are two sides $x, y$ such that
$$
\frac{\sqrt{5}-1}{2} \leq \frac{x}{y} \leq \frac{\sqrt{5}+1}{2} .
$$
$$
\frac{\sqrt{5}-1}{2} \leq \frac{x}{y} \leq \frac{\sqrt{5}+1}{2} .
$$
MathNet 2018-part-05 — 181 problems (EN–VI) Question 82
Math Olympiad Vault · MathNet 2018-part-05 — 181 problems (EN–VI)