[Law of Cosines]

In triangle $ABC$, angle $A$ is $60^{\circ}$, $AB=1$, $BC=a$. Find $AC$.

#
Using the cosine theorem, form a quadratic equation in terms of $A C$ and investigate it.

## Solution

Let $A C=x$. By the cosine theorem from triangle $A B C$, we find that

$$
a^{2}=1+x^{2}-2 x \cos 60^{\circ}=1+x^{2}-x
$$

or

$$
x^{2}-x+1-a^{2}=0
$$

Let $D$ be the discriminant of this quadratic equation. Then $D=4 a^{2}-3$.

If $D>0\left(a>\frac{\sqrt{3}}{2}\right)$, the equation has two distinct solutions: $x=\frac{1 \pm \sqrt{4 a^{2}-3}}{2}$, and if $\frac{\sqrt{3}}{2}<a<1$, both solutions are positive (their product is positive and equal to $1-a^{2}$).

If $a \geqslant 1$, the root $\frac{1+\sqrt{4 a^{2}-3}}{2}$ is positive, while the root $\frac{1-\sqrt{4 a^{2}-3}}{2}$ is negative or 0 (for $a>1$ their product is a negative number $\left.1-a^{2}\right)$.

## Answer

If $a<\frac{\sqrt{3}}{2}$, there are no solutions;

if $a=\frac{\sqrt{3}}{2}, A C=\frac{1}{2}$

if $\frac{\sqrt{3}}{2}<a<1, A C=\frac{1 \pm \sqrt{4 a^{2}-3}}{2}$

if $a \geqslant 1, A C=\frac{1+\sqrt{4 a^{2}-3}}{2}$.