Progetto Olimpiadi di Matematica GARA di SECONDO LIVELLO, 2007

Solution:

The answer is $\mathbf{(E)}$. Letting $a$ be the side of the equilateral triangle, we have the relations $3a = 2(b + h)$, $\frac{a^{2} \sqrt{3}}{4} = \sqrt{3} b h$, from which
$$\left\{ \begin{array}{l} b + h = \frac{3a}{2} \\ bh = \frac{a^{2}}{4} \end{array} \right.$$
The resolving equation of the system, symmetric in the unknowns $b$ and $h$, is $t^{2} - \frac{3a}{2} t + \frac{a^{2}}{4}$, where $t$ is either one of the unknowns. We thus have $4 t^{2} - 6a t + a^{2} = 0$, from which
$$t = \frac{3 \pm \sqrt{9-4}}{4} a$$
from which
$$\frac{b}{h} = \frac{3 + \sqrt{5}}{3 - \sqrt{5}}$$
Rationalizing, we obtain
$$\frac{b}{h} = \frac{7 + 3 \sqrt{5}}{2}$$