$\left.\begin{array}{l}{\left[\begin{array}{l}\text { Law of Cosines } \\ {[\text { Properties and characteristics of a parallelogram }}\end{array}\right]}\end{array}\right]$

Given a parallelogram where the acute angle is $60^{\circ}$. Find the ratio of the sides of the parallelogram if the ratio of the squares of the diagonals is $\frac{1}{3}$.
Express the squares of the diagonals of a parallelogram through its sides using the cosine theorem.

## Solution

Let $d_{1}$ and $d_{2}$ be the diagonals of the parallelogram ($d_{1}<d_{2}$); $a$ and $b$ be the adjacent sides. By the cosine theorem,

$$
\begin{gathered}
d^{2}{ }_{1}=a^{2}+b^{2}-2 a b \cos 60^{\circ}=a^{2}+b^{2}-a b \\
d^{2}{ }_{2}=a^{2}+b^{2}-2 a b \cos 120^{\circ}=a^{2}+b^{2}+a b
\end{gathered}
$$

Then

$$
\frac{d_{1}^{2}}{d_{2}^{2}}=\frac{a^{2}+b^{2}-a b}{a^{2}+b^{2}+a b}=\frac{1}{3}
$$

Therefore,

$$
3 a^{2}+3 b^{2}-3 a b=a^{2}+b^{2}+a b
$$

or

$$
a^{2}+b^{2}-2 a b=0 \text {, or }(a-b)^{2}=0 \text {. }
$$

Thus, $\frac{a}{b}=1$.

## Answer

$1: 1$