HMMT November, 2014

Geometry Difficulty 5.3 AIME, harder Prove it United States

Problem:

In triangle $ABC$ , let the parabola with focus $A$ and directrix $BC$ intersect sides $AB$ and $AC$ at $A_1$ and $A_2$ , respectively. Similarly, let the parabola with focus $B$ and directrix $CA$ intersect sides $BC$ and $BA$ at $B_1$ and $B_2$ , respectively. Finally, let the parabola with focus $C$ and directrix $AB$ intersect sides $CA$ and $CB$ at $C_1$ and $C_2$ , respectively.
If triangle $ABC$ has sides of length $5$ , $12$ , and $13$ , find the area of the triangle determined by lines $A_1C_2$ , $B_1A_2$ , and $C_1B_2$ .

Solution

Solution:

By the definition of a parabola, we get $AA_1 = A_1B \sin B$ and similarly for the other points. So $\frac{AB_2}{AB} = \frac{AC_1}{AC}$ , giving $B_2C_1 \parallel BC$ , and similarly for the other sides. So $DEF$ (WLOG, in that order) is similar to $ABC$ . It suffices to scale after finding the length of $EF$ , which is

$\frac{EF}{BC} = \frac{2 \prod_{cyc} \sin A + \sum_{cyc} \sin A \sin B - 1}{\prod_{cyc}(1+\sin A)}$

Plugging in, squaring the result, and multiplying by $K_{ABC} = 30$ gives the answer.

$\boxed{\dfrac{6728}{3375}}$

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