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Geometry Difficulty 5.3 AIME, harder Prove it United States

Problem:

Let $\omega$ be a circle, and let $ABCD$ be a quadrilateral inscribed in $\omega$ . Suppose that $BD$ and $AC$ intersect at a point $E$ . The tangent to $\omega$ at $B$ meets line $AC$ at a point $F$ , so that $C$ lies between $E$ and $F$ . Given that $AE = 6$ , $EC = 4$ , $BE = 2$ , and $BF = 12$ , find $DA$ .

Solution

Solution:

Answer: $2 \sqrt{42}$

By power of a point, we have $ED \cdot EB = EA \cdot EC$ , whence $ED = 12$ .

Additionally, by power of a point, we have $144 = FB^2 = FC \cdot FA = FC(FC + 10)$ , so $FC = 8$ .

Note that $\angle FBC = \angle FAB$ and $\angle CFB = \angle AFB$ , so $\triangle FBC \sim \triangle FAB$ . Thus, $AB / BC = FA / FB = 18 / 12 = 3 / 2$ , so $AB = 3k$ and $BC = 2k$ for some $k$ .

Since $\triangle BEC \sim \triangle AED$ , we have $AD / BC = AE / BE = 3$ , so $AD = 3 BC = 6k$ .

By Stewart's theorem on $\triangle EBF$ , we have

$(4)(8)(12) + (2k)^2(12) = (2)^2(8) + (12)^2(4)$

whence $8 + k^2 = 8 / 12 + 12$ .

Thus, $k^2 = 14 / 3$ . Therefore,
$DA = 6k = 6 \sqrt{14 / 3} = 6 \frac{\sqrt{42}}{3} = 2 \sqrt{42}.$

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.