2003 AIME II Problems/Problem 14

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

The y-coordinate of $F$ must be $4$ . All other cases yield non-convex hexagons, which violate the problem statement.

Letting $F = (f,4)$ , and knowing that $\angle FAB = 120^\circ$ , we can use rewrite $F$ using complex numbers: $f + 4 i = (b + 2 i)\left(e^{(2 \pi / 3)}\right) = (b + 2 i)\left(-1/2 + \frac{\sqrt{3}}{2} i\right) = -\frac{b}{2}-\sqrt{3}+\left(\frac{b\sqrt{3}}{2}-1\right)i$ . We solve for $b$ and $f$ and find that $F = \left(\frac{8}{\sqrt{3}}, 4\right)$ and that $B = \left(\frac{10}{\sqrt{3}}, 2\right)$ .

The area of the hexagon can then be found as the sum of the areas of two congruent triangles ( $EFA$ and $BCD$ , with height $8$ and base $\frac{8}{\sqrt{3}}$ and a parallelogram ( $ABDE$ , with height $8$ and base $\frac{10}{\sqrt{3}}$ .

$A = 2 \times \frac{1}{2} \times 8 \times \frac{8}{\sqrt{3}} + 8 \times \frac{10}{\sqrt{3}} = \frac{144}{\sqrt{3}} = 48\sqrt{3}.

Thus,$ (Error compiling LaTeX. Unknown error_msg)m+n = \boxed{51}$.

== Solution (Incomplete/incorrect) == {{image}}

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

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

<center>$ (Error compiling LaTeX. Unknown error_msg)\sin{60}\cos{x}-\cos{60}\sin{x}=\frac{\sqrt{3}\cos{x}}{2}-\frac{\sin{x}}{2}=2\sin{x}$</center>

Isolating$ (Error compiling LaTeX. Unknown error_msg)\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 following is incorrect as the hexagon is NOT regular (although it is equilateral). The previous work IS correct, so I am leaving it as part of an incomplete solution'''

The area of the hexagon is$ (Error compiling LaTeX. Unknown error_msg)\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 • Answer Key • Resources)
Preceded by Problem 13	Followed by Problem 15
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

Page

Toolbox

Search

2003 AIME II Problems/Problem 14

Problem

Solution

See also