Difference between revisions of "2003 AIME II Problems/Problem 14"
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Letting <math>F = (f,4)</math>, and knowing that <math>\angle FAB = 120^\circ</math>, we can use rewrite <math>F</math> using complex numbers: | Letting <math>F = (f,4)</math>, and knowing that <math>\angle FAB = 120^\circ</math>, we can use rewrite <math>F</math> using complex numbers: | ||
| − | <math>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 < | + | <math>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</math>. We solve for <math>b</math> and <math>f</math> and find that <math>F = \left(\frac{8}{\sqrt{3}}, 4\right)</math> and that <math>B = \left(\frac{10}{\sqrt{3}}, 2\right)</math>. |
| − | The area of the hexagon can then be found as the sum of the areas of two congruent triangles (< | + | The area of the hexagon can then be found as the sum of the areas of two congruent triangles (<math>EFA</math> and <math>BCD</math>, with height <math>8</math> and base <math>\frac{8}{\sqrt{3}}</math> and a parallelogram (<math>ABDE</math>, with height <math>8</math> and base <math>\frac{10}{\sqrt{3}}</math>. |
| − | < | + | <math>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, <math>m+n = \boxed{51}< | + | Thus, </math>m+n = \boxed{51}<math>. |
== Solution (Incomplete/incorrect) == | == Solution (Incomplete/incorrect) == | ||
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{{image}} | {{image}} | ||
| − | In this image, we have drawn perpendiculars to the <math>x< | + | In this image, we have drawn perpendiculars to the </math>x<math>-axis from F and B, and have labeled the angle between the </math>x<math>-axis and segment </math>AB<math> </math>x<math>. Thus, the angle between the </math>x<math>-axis and segment </math>AF<math> is </math>60-x<math> so, </math>\sin{(60-x)}=2\sin{x}<math>. Expanding, we have |
| − | <center><math>\sin{60}\cos{x}-\cos{60}\sin{x}=\frac{\sqrt{3}\cos{x}}{2}-\frac{\sin{x}}{2}=2\sin{x}< | + | <center></math>\sin{60}\cos{x}-\cos{60}\sin{x}=\frac{\sqrt{3}\cos{x}}{2}-\frac{\sin{x}}{2}=2\sin{x}<math></center> |
| − | Isolating <math>\sin{x}< | + | Isolating </math>\sin{x}<math> we see that </math>\frac{\sqrt{3}\cos{x}}{2}=\frac{5\sin{x}}{2}<math>, or </math>\cos{x}=\frac{5}{\sqrt{3}}\sin{x}<math>. Using the fact that </math>\sin^2{x}+\cos^2{x}=1<math>, we have </math>\frac{28}{3}\sin^2{x}=1<math>, or </math>\sin{x}=\sqrt{\frac{3}{28}}<math>. Letting the side length of the hexagon be </math>y<math>, we have </math>\frac{2}{y}=\sqrt{\frac{3}{28}}<math>. After simplification we see that </math>y=\frac{4\sqrt{21}}{3}<math>. |
'''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 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 <math>\frac{3y^2\sqrt{3}}{2}< | + | The area of the hexagon is </math>\frac{3y^2\sqrt{3}}{2}<math>, so the area of the hexagon is </math>\frac{3*\frac{16*21}{3^2}*\sqrt{3}}{2}=56\sqrt{3}<math>, or </math>m+n=\boxed{059}$. |
== See also == | == See also == | ||
{{AIME box|year=2003|n=II|num-b=13|num-a=15}} | {{AIME box|year=2003|n=II|num-b=13|num-a=15}} | ||
[[Category:Intermediate Geometry Problems]] | [[Category:Intermediate Geometry Problems]] | ||
Revision as of 16:28, 5 January 2010
Problem
Let 
and 
be points on the coordinate plane. Let 
be a convex equilateral hexagon such that 
and the y-coordinates of its vertices are distinct elements of the set 
The area of the hexagon can be written in the form 
where 
and 
are positive integers and n is not divisible by the square of any prime. Find 
Solution
The y-coordinate of 
must be 
. All other cases yield non-convex hexagons, which violate the problem statement.
Letting 
, and knowing that 
, we can use rewrite using complex numbers:
. We solve for 
and 
and find that 
and that 
.
The area of the hexagon can then be found as the sum of the areas of two congruent triangles (
and 
, with height 
and base 
and a parallelogram (
, with height and base 
.
$A = 2 \times \frac{1}{2} \times 8 \times \frac{8}{\sqrt{3}} + 8 \times \frac{10}{\sqrt{3}} = \frac{144}{\sqrt{3}} = 48\sqrt{3}.
Thus,$ (Error compiling LaTeX. Unknown error_msg)m+n = \boxed{51}$.
== Solution (Incomplete/incorrect) == {{image}}
From this image, we can see that the y-coordinate of F is 4, and from this, we can gather that the coordinates of E, D, and C, respectively, are 8, 10, and 6.
{{image}}
In this image, we have drawn perpendiculars to the$ (Error compiling LaTeX. Unknown error_msg)x
x
AB$$ (Error compiling LaTeX. Unknown error_msg)x
xAF
60-x
\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}
\frac{\sqrt{3}\cos{x}}{2}=\frac{5\sin{x}}{2}
\cos{x}=\frac{5}{\sqrt{3}}\sin{x}
\sin^2{x}+\cos^2{x}=1
\frac{28}{3}\sin^2{x}=1\sin{x}=\sqrt{\frac{3}{28}}
y\frac{2}{y}=\sqrt{\frac{3}{28}}
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}
\frac{3*\frac{16*21}{3^2}*\sqrt{3}}{2}=56\sqrt{3}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 | ||
