Problem 3. The lengths of the sides of a triangle are in a geometric progression with a common ratio $r$. Find the values of $r$ for which the triangle is, respectively, acute, right, or obtuse.
## Solution;

We can assume $r \geq 1$, since if not, we can simply reverse the order in which the sides are considered. If we take the length of the shortest side as the unit of length, then the lengths are $1, r$, and $r^{2}$. Let $a$ be the angle opposite to $r^{2}$, which is the largest. By the law of cosines, we have:

$$
r^{4}=1+r^{2}-2 r \cos a
$$

The triangle is right-angled if and only if $\cos a=0$, that is, $r^{4}-r^{2}-1=0$.

In other words: $r^{2}=\frac{1+\sqrt{5}}{2}$. And, ultimately, $r=\sqrt{\frac{1+\sqrt{5}}{2}}$.

It is now immediate to deduce that the triangle is acute ( $\cos a>0$ ) if and only if

$$
1 \leq r<\sqrt{\frac{1+\sqrt{5}}{2}}
$$

Similarly, the triangle is obtuse ( $\cos a<0$ ) if and only if $\sqrt{\frac{1+\sqrt{5}}{2}}<r<\frac{1+\sqrt{5}}{2}$

The last inequality is forced by the fact that for the triangle to exist, it must be:

$$
r^{2}<1+r
$$