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2003 AIME II Problems/Problem 14

Revision as of 15:31, 5 January 2010 by Haoye (talk | contribs) (→‎Solution)

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 and/or degenerate 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: $f + 4 i = (b + 2 i)\left(e^{i(2 \pi / 3)}\right) = (b + 2 i)\left(-1/2 + \frac{\sqrt{3}}{2} i\right) = -\frac{b}{2}-\sqrt{3}+\left(\frac{b\sqrt{3}}{2}-1\right)i$. 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, $m+n = \boxed{51}$.

Solution (Incomplete/incorrect)

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

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.

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 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 $\frac{3y^2\sqrt{3}}{2}$, so the area of the hexagon is $\frac{3*\frac{16*21}{3^2}*\sqrt{3}}{2}=56\sqrt{3}$, or $m+n=\boxed{059}$.

See also

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
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