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"
(→‎Solution (Incomplete/incorrect))
Line 15: Line 15:
 
Thus, <math>m+n = \boxed{051}</math>.
 
Thus, <math>m+n = \boxed{051}</math>.
  
−
== Solution (Incomplete/incorrect) ==
+
== Solution 2 ==
 
<asy>
 
<asy>
 
size(200);
 
size(200);
Line 24: Line 24:
 
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.
 
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}}
+
<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>
  
−
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
+
Let the angle between the <math>x</math>-axis and segment <math>AB</math> be  <math>\theta</math>, as shown above. Thus, as <math>\angle FAB=120^\circ</math>, the angle between the <math>x</math>-axis and segment <math>AF</math> is <math>60-\theta</math>, so <math>\sin{(60-\theta)}=2\sin{\theta}</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}</math></center>
+
<center><math>\sin{60}\cos{\theta}-\cos{60}\sin{\theta}=\frac{\sqrt{3}\cos{\theta}}{2}-\frac{\sin{\theta}}{2}=2\sin{\theta}</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>.  
+
Isolating <math>\sin{\theta}</math> we see that <math>\frac{\sqrt{3}\cos{\theta}}{2}=\frac{5\sin{\theta}}{2}</math>, or <math>\cos{\theta}=\frac{5}{\sqrt{3}}\sin{\theta}</math>. Using the fact that <math>\sin^2{\theta}+\cos^2{\theta}=1</math>, we have <math>\frac{28}{3}\sin^2{\theta}=1</math>, or <math>\sin{\theta}=\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 find that 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'''
+
In particular, note that by the Pythagorean theorem, <math>b^2+2^2=y^2</math>, hence <math>b=10\sqrt{3}/3</math>. Also, if <math>C=(c,6)</math>, then <math>y^2=BC^2=4^2+(c-b)^2</math>, hence <math>c-b=8\sqrt{3}/3,</math> and thus <math>c=18\sqrt{3}/3</math>. Using similar methods (or symmetry), we determine that <math>D=(10\sqrt{3}/3,10)</math>, <math>E=(0,8)</math>, and <math>F=(-8\sqrt{3}/3,4)</math>. By the Shoelace theorem,
 +
<cmath>[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}.</cmath>
  
−
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>.
+
Hence the answer is <math>\boxed{51}</math>.
  
 
== See also ==
 
== See also ==

Revision as of 19:53, 13 March 2015

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

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 (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

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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2003_AIME_II_Problems/Problem_14&oldid=68847"