Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2003 AIME II Problems/Problem 14"
Line 3: Line 3:
  
 
== Solution ==
 
== Solution ==
 +
 +
The y-coordinate of <math>F</math> must be <math>4</math>.  All other cases yield non-convex hexagons, which violate the problem statement.
 +
 +
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>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 (</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}</math>.
 +
 +
== Solution (Incomplete/incorrect) ==
 
{{image}}
 
{{image}}
  
−
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.
+
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}}
 
{{image}}
Line 13: Line 26:
 
<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>
 
<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}</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 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}</math>.
+
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 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}</math>.
  
 
== 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 15:28, 5 January 2010

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 ($ (Error compiling LaTeX. Unknown error_msg)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
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2003_AIME_II_Problems/Problem_14&oldid=33157"