2003 AIME II Problems/Problem 14

Revision as of 19:53, 13 March 2015 by Mathgeek2006 (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 1

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, $m+n = \boxed{051}$.

Solution 2

[asy] size(200); draw((0,0)--(10/sqrt(3),2)--(18/sqrt(3),6)--(10/sqrt(3),10)--(0,8)--(-8/sqrt(3),4)--cycle); dot((0,0));dot((10/sqrt(3),2));dot((18/sqrt(3),6));dot((10/sqrt(3),10));dot((0,8));dot((-8/sqrt(3),4)); label("$A (0,0)$",(0,0),S);label("$B (b,2)$",(10/sqrt(3),2),SE);label("$C$",(18/sqrt(3),6),E);label("$D$",(10/sqrt(3),10),N);label("$E$",(0,8),NW);label("$F$",(-8/sqrt(3),4),W); [/asy] 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.

[asy] size(200); draw((0,0)--(10/sqrt(3),2)--(18/sqrt(3),6)--(10/sqrt(3),10)--(0,8)--(-8/sqrt(3),4)--cycle); dot((0,0));dot((10/sqrt(3),2));dot((18/sqrt(3),6));dot((10/sqrt(3),10));dot((0,8));dot((-8/sqrt(3),4)); label("$A (0,0)$",(0,0),SE);label("$B (b,2)$",(10/sqrt(3),2),SE);label("$C$",(18/sqrt(3),6),E);label("$D$",(10/sqrt(3),10),N);label("$E$",(0,8),NW);label("$F$",(-8/sqrt(3),4),W); xaxis("$x$");yaxis("$y$"); pair b=foot((10/sqrt(3),2),(0,0),(10,0)); pair f=foot((-8/sqrt(3),4),(0,0),(-10,0)); draw(b--(10/sqrt(3),2),dotted); draw(f--(-8/sqrt(3),4),dotted); label("$\theta$",(0,0),7*dir((0,0)--(10/sqrt(3),2)+(4*sqrt(21)/3,0))); [/asy]

Let the angle between the $x$-axis and segment $AB$ be $\theta$, as shown above. Thus, as $\angle FAB=120^\circ$, the angle between the $x$-axis and segment $AF$ is $60-\theta$, so $\sin{(60-\theta)}=2\sin{\theta}$. Expanding, we have

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

Isolating $\sin{\theta}$ we see that $\frac{\sqrt{3}\cos{\theta}}{2}=\frac{5\sin{\theta}}{2}$, or $\cos{\theta}=\frac{5}{\sqrt{3}}\sin{\theta}$. Using the fact that $\sin^2{\theta}+\cos^2{\theta}=1$, we have $\frac{28}{3}\sin^2{\theta}=1$, or $\sin{\theta}=\sqrt{\frac{3}{28}}$. Letting the side length of the hexagon be $y$, we have $\frac{2}{y}=\sqrt{\frac{3}{28}}$. After simplification we find that that $y=\frac{4\sqrt{21}}{3}$.

In particular, note that by the Pythagorean theorem, $b^2+2^2=y^2$, hence $b=10\sqrt{3}/3$. Also, if $C=(c,6)$, then $y^2=BC^2=4^2+(c-b)^2$, hence $c-b=8\sqrt{3}/3,$ and thus $c=18\sqrt{3}/3$. Using similar methods (or symmetry), we determine that $D=(10\sqrt{3}/3,10)$, $E=(0,8)$, and $F=(-8\sqrt{3}/3,4)$. By the Shoelace theorem, \[[ABCDEF]=\frac12\left|\begin{array}{cc} 0&0\\ 10\sqrt{3}/3&2\\ 18\sqrt{3}/3&6\\ 10\sqrt{3}/3&10\\ 0&8\\ -8\sqrt{3}/3&4\\ 0&0\\ \end{array}\right|=\frac12|60+180+80-36-60-(-64)|\sqrt{3}/3=48\sqrt{3}.\]

Hence the answer is $\boxed{51}$.

See also

2003 AIME II (ProblemsAnswer KeyResources)
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Invalid username
Login to AoPS