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

2003 AIME II Problems/Problem 14

Revision as of 15:14, 19 February 2020 by Ant08 (talk | contribs) (→‎Solution 4 (No Trig))

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 $\boxed{51}$.

Solution 3

This is similar to solution 2 but faster and easier. First off we see that the y coordinate of F must be 4, the y coordinate of E must be 8, the y coordinate of D must be 10, and the y coordinate of C must be 6 (from the parallel sides of the hexagon). We then use the sine sum angle formula to find the x coordinate of B (lets call it $x$): $2\cdot\cos(120)+x\sin(120)=\frac{x\sqrt3}{2}-1=4\rightarrow x=\frac{10\sqrt3}{3}$. Now that we know $x$ we can find the x coordinate of F in multiple ways, including using the cosine sum angle formula or using the fact that triangle AFE is isosceles and AE is on the y axis. Either way, we find that the x coordinate of F is $-\frac{8\sqrt3}{3}$. Now, divide ABCDEF into two congruent triangles and a parallelogram: AFE, BCD, and ABDE. The areas of AFE and BCD are each $\frac12\cdot\frac{8\sqrt3}{3}\cdot8=\frac{32\sqrt3}{3}$. The area of ABDE is $\frac12\cdot8\cdot\frac{10\sqrt3}{3}=\frac{80\sqrt3}3$. The total area of the hexagon is $2\cdot\frac{32\sqrt3}3+\frac{80\sqrt3}3=\frac{144\sqrt3}{3}=48\sqrt3\rightarrow48+3=\boxed{051}$

Solution 4 (No Trig)

This is also similar to solution 2 but does not use any trig.

First, we see that the y-coordinates of F, E, D, and C must be 4, 8, and 10, and 6, respectively, as in the previous solution. We can draw a rectangle around hexagon ABCDEF and use negative space to find the area of the hexagon. If we call the distance from the foot of the perpendiculars of B and F to A $x$ and $z$, respectively, and the distance from the bottom left vertex of the rectangle to the foot of the perpendicular from B $y$. This tells us that the area of the entire rectangle is $10(x+y+z)$. Then, if we find the area of the extra triangles and subtract, we find that the area of hexagon ABCDEF as $6x+8z+2y$. However, noticing that $x=y$, the area of ABCDEF can also be expressed as $8(x+z)$. Now we just need to find $x+z$. Since $AB=AF$ and $\angle BAF = 120$ degrees, $BF=AB\sqrt{3}$. However, we can find AB by using the Pythagorean Theorem on either of the right triangles formed by dropping perpendiculars from B and F to the x-axis (let's call them ABX and AFY). From triangle ABX we have that $AB=\sqrt{4+x^2}$, so $BF=\sqrt{3x^2+12}$. Since AB=AF, we can also form the equation $4+x^2=16+z^2$. We can also find BF by dropping a perpendicular from B to line FY and using the Pythagorean Theorem on the right triangle formed. This gives us $BF=\sqrt{4+(x+z)^2}$. Setting our two values of BF equal and substituting $x^2$ as $12+z^2$ and simplifying, we get the equation $3z^4-16z^2-1024=0$. Now we can use the quadratic formula to get that $z^2=\frac{64}{3}$ or -18, so $z^2=\frac{64}{3}$. Plugging this value back into the equation $x^2=12+z^2$, we get that $x^2=\frac{100}{3}$. Now we get that $x+z$ is $6\sqrt{3}$, so the area of the hexagon is $8 \cdot 6\sqrt{3}=48\sqrt{3}$, so the answer is $48+3=\boxed{051}$

~ant08 and sky2025

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=118123"