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Geometry Difficulty 5.3 AIME, harder Prove it United States

Problem:

Let $ABCD$ be a parallelogram with $AB = 8$ , $AD = 11$ , and $\angle BAD = 60^{\circ}$ . Let $X$ be on segment $CD$ with $CX / XD = 1 / 3$ and $Y$ be on segment $AD$ with $AY / YD = 1 / 2$ . Let $Z$ be on segment $AB$ such that $AX$ , $BY$ , and $DZ$ are concurrent. Determine the area of triangle $XYZ$ .

Solution

Solution:

Answer: $\dfrac{19 \sqrt{3}}{2}$

Let $AX$ and $BD$ meet at $P$ . We have $DP / PB = DX / AB = 3 / 4$ . Now, applying Ceva's Theorem in triangle $ABD$ , we see that
$\frac{AZ}{ZB} = \frac{DP}{PB} \cdot \frac{AY}{YD} = \frac{3}{4} \cdot \frac{1}{2} = \frac{3}{8}$
Now,
$\frac{[AYZ]}{[ABCD]} = \frac{[AYZ]}{2[ABD]} = \frac{1}{2} \cdot \frac{1}{3} \cdot \frac{3}{11} = \frac{1}{22},$
and similarly
$\frac{[DYX]}{[ABCD]} = \frac{1}{2} \cdot \frac{2}{3} \cdot \frac{3}{4} = \frac{1}{4}.$
Also,
$\frac{[XCBZ]}{[ABCD]} = \frac{1}{2}\left(\frac{1}{4} + \frac{8}{11}\right) = \frac{43}{88}$
The area of $XYZ$ is the rest of the fraction of the area of $ABCD$ not covered by the three above polygons, which by a straightforward calculation is $19 / 88$ the area of $ABCD$ , so our answer is
$8 \cdot 11 \cdot \sin 60^{\circ} \cdot \frac{19}{88} = \frac{19 \sqrt{3}}{2}$

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