HMMT November, 2018

Geometry Difficulty 5.1 AIME, harder Prove it United States

Problem:

Points $E, F, G, H$ are chosen on segments $AB, BC, CD, DA$ , respectively, of square $ABCD$ . Given that segment $EG$ has length $7$ , segment $FH$ has length $8$ , and that $EG$ and $FH$ intersect inside $ABCD$ at an acute angle of $30^{\circ}$ , then compute the area of square $ABCD$ .

Solution

Solution:

Rotate $EG$ by $90^{\circ}$ about the center of the square to $E'G'$ with $E' \in AD$ and $G' \in BC$ . Now $E'G'$ and $FH$ intersect at an angle of $60^{\circ}$ . Then consider the translation which takes $E'$ to $H$ and $G'$ to $I$ . Triangle $FHI$ has $FH = 8$ , $HI = 7$ and $\angle FHI = 60^{\circ}$ . Furthermore, the height of this triangle is the side length of the square. Using the Law of Cosines,
$FI = \sqrt{7^2 - 7 \cdot 8 + 8^2} = \sqrt{57}$
By computing the area of $FHI$ in two ways, if $h$ is the height then
$\frac{1}{2} \times \sqrt{57} \times h = \frac{1}{2} \times \frac{\sqrt{3}}{2} \times 7 \times 8$
Then $h = \frac{28}{\sqrt{19}}$ and the area of the square is $h^2 = \frac{784}{19}$ .

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