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Weekly Problem 15 - 2006

Stage: 4 Challenge Level: Challenge Level:3 Challenge Level:3 Challenge Level:3

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.

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