Let the widescreen width and traditional width be $w$ and $W$ respectively. Then the respective heights are $\frac{9w}{16}$ and $\frac{3W}{4}. $ As the areas are equal: $$x \times \frac{9w}{16} = W \times \frac{2W}{4}$$ i.e. $$w^2 = \frac {4}{3}W^2$$ Hence $w : W = 2 : \sqrt{3}$.

*This problem is taken from the UKMT Mathematical Challenges.*