Difference between revisions of "2003 AIME II Problems/Problem 14"
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Hence the answer is <math>\boxed{51}</math>. | Hence the answer is <math>\boxed{51}</math>. | ||
+ | ===Note=== | ||
+ | By symmetry the area of <math>ABCDEF</math> is twice the area of <math>ABCF</math>. Therefore, you only need to calculate the coordinates of <math>B</math>, <math>C</math>, and <math>F</math>. | ||
== Solution 3 == | == Solution 3 == | ||
This is similar to solution 2 but faster and easier. | This is similar to solution 2 but faster and easier. | ||
Line 65: | Line 67: | ||
==Solution 4 (No Trig)== | ==Solution 4 (No 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 | + | <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> |
+ | |||
+ | 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 solutions. We can draw a rectangle around the 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 <math>x</math> and <math>z</math>, respectively, and the distance from the bottom left vertex of the rectangle to the foot of the perpendicular from B <math>y</math>. This tells us that the area of the entire rectangle is <math>10(x+y+z)</math>, since the opposite sides are parallel and thus the length of the rectangle is <math>4+4+2=10</math>. Then, | ||
+ | if we find the area of the extra triangles and subtract, we find that the area of hexagon ABCDEF as <math>6x+8z+2y</math>. However, noticing that <math>x=y</math>, the area of ABCDEF can also be expressed as <math>8(x+z)</math>. Now we just need to find <math>x+z</math>. Since <math>AB=AF</math> and <math>\angle BAF = 120</math> degrees, <math>BF=AB\sqrt{3}</math>. 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 <math>AB=\sqrt{4+x^2}</math>, so <math>BF=\sqrt{3x^2+12}</math>. Since AB=AF, we can also form the equation <math>4+x^2=16+z^2</math>. | From triangle ABX we have that <math>AB=\sqrt{4+x^2}</math>, so <math>BF=\sqrt{3x^2+12}</math>. Since AB=AF, we can also form the equation <math>4+x^2=16+z^2</math>. | ||
− | 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 <math>BF=\sqrt{4+(x+z)^2}</math>. Setting our two values of BF equal and substituting <math>x^2</math> as <math>12+z^2</math> and simplifying, we get the equation <math>3z^4-16z^2-1024=0</math>. Now we can use the quadratic formula to get that <math>z^2=\frac{64}{3}</math> or -18, so <math>z^2=\frac{64}{3}</math>. Plugging this value back into the equation <math>x^2=12+z^2</math>, we get that <math>x^2=\frac{100}{3}</math>. Now we get that x+z is <math>6\sqrt{3}</math>, so the area of the hexagon is <math>8 \cdot 6\sqrt{3}=48\sqrt{3}</math>, so the answer is <math>48+3=\boxed{051}</math> | + | 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 <math>BF=\sqrt{4+(x+z)^2}</math>. Setting our two values of BF equal and substituting <math>x^2</math> as <math>12+z^2</math> and simplifying, we get the equation <math>3z^4-16z^2-1024=0</math>. Now we can use the quadratic formula to get that <math>z^2=\frac{64}{3}</math> or <math>-18</math>, so <math>z^2=\frac{64}{3}</math>. Plugging this value back into the equation <math>x^2=12+z^2</math>, we get that <math>x^2=\frac{100}{3}</math>. Now we get that <math>x+z</math> is <math>6\sqrt{3}</math>, so the area of the hexagon is <math>8 \cdot 6\sqrt{3}=48\sqrt{3}</math>, so the answer is <math>48+3=\boxed{051}</math> |
~ant08 and sky2025 | ~ant08 and sky2025 | ||
− | + | ||
+ | ==Video Solution by Sal Khan== | ||
+ | https://www.youtube.com/watch?v=Ec-BKdC8vOo&list=PLSQl0a2vh4HCtW1EiNlfW_YoNAA38D0l4&index=4 | ||
+ | - AMBRIGGS | ||
+ | |||
== 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]] | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 15:09, 18 August 2023
Contents
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 1
The y-coordinate of must be
. All other cases yield non-convex and/or degenerate 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
).
.
Thus, .
Solution 2
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.
Let the angle between the -axis and segment
be
, as shown above. Thus, as
, the angle between the
-axis and segment
is
, so
. Expanding, we have
![$\sin{60}\cos{\theta}-\cos{60}\sin{\theta}=\frac{\sqrt{3}\cos{\theta}}{2}-\frac{\sin{\theta}}{2}=2\sin{\theta}$](http://latex.artofproblemsolving.com/5/4/a/54a8d083055a7fdb510e38757f98ced666d872db.png)
Isolating we see that
, or
. Using the fact that
, we have
, or
. Letting the side length of the hexagon be
, we have
. After simplification we find that that
.
In particular, note that by the Pythagorean theorem, , hence
. Also, if
, then
, hence
and thus
. Using similar methods (or symmetry), we determine that
,
, and
. By the Shoelace theorem,
Hence the answer is .
Note
By symmetry the area of is twice the area of
. Therefore, you only need to calculate the coordinates of
,
, and
.
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 ):
.
Now that we know
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
.
Now, divide ABCDEF into two congruent triangles and a parallelogram: AFE, BCD, and ABDE. The areas of AFE and BCD are each
. The area of ABDE is
.
The total area of the hexagon is
Solution 4 (No 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 solutions. We can draw a rectangle around the 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 and
, respectively, and the distance from the bottom left vertex of the rectangle to the foot of the perpendicular from B
. This tells us that the area of the entire rectangle is
, since the opposite sides are parallel and thus the length of the rectangle is
. Then,
if we find the area of the extra triangles and subtract, we find that the area of hexagon ABCDEF as
. However, noticing that
, the area of ABCDEF can also be expressed as
. Now we just need to find
. Since
and
degrees,
. 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
, so
. Since AB=AF, we can also form the equation
.
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
. Setting our two values of BF equal and substituting
as
and simplifying, we get the equation
. Now we can use the quadratic formula to get that
or
, so
. Plugging this value back into the equation
, we get that
. Now we get that
is
, so the area of the hexagon is
, so the answer is
~ant08 and sky2025
Video Solution by Sal Khan
https://www.youtube.com/watch?v=Ec-BKdC8vOo&list=PLSQl0a2vh4HCtW1EiNlfW_YoNAA38D0l4&index=4 - AMBRIGGS
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.