The sides of a triangle form a geometric progression. What values can the ratio of the progression take?






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If the sides of a triangle form a geometric progression, the ratio of the progression - depending on whether we start from the smallest or the largest side - is greater than or equal to 1 or less than or equal to 1; these two values are reciprocals of each other.

We can therefore assume that the ratio of the progression is $q \geq 1$. (In the case of $q=1$, we have an equilateral triangle). The sides of the triangle are: $a, a q, a q^{2}$. The condition for these three numbers to actually be the lengths of the sides of a triangle is:

$$
a q^{2}<a q+a \quad \text { or } \quad q^{2}<q+1 \quad \text { or } \quad q^{2}-q-1<0
$$

This inequality is satisfied if $q$ lies between the roots of the equation

$$
q^{2}-q-1=0
$$

These roots are of opposite signs. The positive root is

$$
q_{1}=\frac{1+\sqrt{5}}{2}
$$

Since $q<q_{1}$, but according to the previous conditions $q \geq 1$, the inequality $q^{2}-q-1<0$ is satisfied if

$$
1 \leq q \leq \frac{\sqrt{5}+1}{2}
$$

However, as we have seen, $q$ can also take the reciprocals of the values within this range. Since

$$
\frac{2}{\sqrt{5}+1}=\frac{2(\sqrt{5}-1)}{5-1}=\frac{\sqrt{5}-1}{2}
$$

it is clearly true that:

$$
\frac{\sqrt{5}-1}{2}<q<\frac{\sqrt{5}+1}{2}
$$

Vásárhelyi Nagy Sándor (Kegyesrendi rg. VI. o. Sátoraljaújhely)[^0] such a function is negative between the zeros.

[^0]:    ${ }^{1} f(q) \equiv q^{2}-q-1$ is a quadratic function of $q$ where the coefficient of the quadratic term is positive and it has real zeros.