2003 AIME II Problems/Problem 14

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Problem

Let $A = (0,0)$ and $B = (b,2)$ be points on the coordinate plane. Let $ABCDEF$ be a convex equilateral hexagon such that $\angle FAB = 120^\circ,$ $\overline{AB}\parallel \overline{DE},$ $\overline{BC}\parallel \overline{EF,}$ $\overline{CD}\parallel \overline{FA},$ and the y-coordinates of its vertices are distinct elements of the set $\{0,2,4,6,8,10\}.$ The area of the hexagon can be written in the form $m\sqrt {n},$ where $m$ and $n$ are positive integers and n is not divisible by the square of any prime. Find $m + n.$

Solution

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From this image, we can see that the oordinate of F is 4, and from this, we can gather that the oordinates of E, D, and C, respectively, are 8, 10, and 6.

An image is supposed to go here. You can help us out by creating one and editing it in. Thanks.

In this image, we have drawn perpendiculars to the $x$ -axis from F and B, and have labeled the angle between the $x$ -axis and segment $AB$ $x$ . Thus, the angle between the $x$ -axis and segment $AF$ is $60-x$ so, $\sin{(60-x)}=2\sin{x}$ . Expanding, we have

$\sin{60}\cos{x}-\cos{60}\sin{x}=\frac{\sqrt{3}\cos{x}}{2}-\frac{\sin{x}}{2}=2\sin{x}$

Isolating $\sin{x}$ we see that $\frac{\sqrt{3}\cos{x}}{2}=\frac{5\sin{x}}{2}$ , or $\cos{x}=\frac{5}{\sqrt{3}}\sin{x}$ . Using the fact that $\sin^2{x}+\cos^2{x}=1$ , we have $\frac{28}{3}\sin^2{x}=1$ , or $\sin{x}=\sqrt{\frac{3}{28}}$ . Letting the side length of the hexagon be $y$ , we have $\frac{2}{y}=\sqrt{\frac{3}{28}}$ . After simplification we see that $y=\frac{4\sqrt{21}}{3}$ . The area of the hexagon is $\frac{3y^2\sqrt{3}}{2}$ , so the area of the hexagon is $\frac{3*\frac{16*21}{3^2}*\sqrt{3}}{2}=56\sqrt{3}$ , or $m+n=\boxed{059}$ .

2003 AIME II (Problems • Resources)
Preceded by Problem 13	Followed by Problem 15
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

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2003 AIME II Problems/Problem 14

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Solution

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