2017/18 AHSMC Part 1 Problem 10

Problem 10 of 16

In a trapezoid $ABCD$ , with $AD$ parallel to $BC$ the diagonals intersect at $O$ . If the area of $\triangle AOD$ is $\dfrac{9}{16}$ of the area of the trapezoid then the ratio of the areas of triangles $AOD$ to $BOC$ is equal to

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Solution

Let $\mathrm{area}(AOD) = a$ , $\mathrm{area}(BOC) = b$ , $\mathrm{area}(AOB) = \mathrm{area}(COD) = c$ . If $\dfrac{a}{b} = x$ then $\dfrac{a}{c} = \dfrac{AO}{OC} = \dfrac{c}{b}$ , so $\dfrac{a}{c} = \dfrac{c}{b} = \sqrt{x}$ , hence $\frac{\mathrm{area}(AOD)}{\mathrm{area}(ABCD)} = \frac{a}{a + b + 2c} = \frac{\frac{a}{b}}{2\frac{c}{b} + \frac{a}{b} + 1} = \frac{x}{(\sqrt{x} + 1)^2} = \frac{9}{16}.$ Solving for $x$ one obtains $x = 9$ . The answer is (C).

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