Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 5.0 AIME, harder Find the answer

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 $DA$ .

Your answer

A number or a short expression. Spacing and $ signs are ignored.

Solution

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 $AD=3BC=6k$ . By Stewart's theorem on $\triangle EBF$ , we have $(4)(8)(12)+(2k)^{2}(12)=(2)^{2}(8)+(12)^{2}(4) \Longrightarrow 8+k^{2}=8 / 12+12$ whence $k^{2}=14 / 3$ . Thus, $DA=6k=6 \sqrt{14 / 3}=6 \frac{\sqrt{42}}{3}=2 \sqrt{42}$ .

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