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>. | ||
+ | |||
+ | == 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 <math>x</math>): <math>2\cdot\cos(120)+x\sin(120)=\frac{x\sqrt3}{2}-1=4\rightarrow x=\frac{10\sqrt3}{3}</math>. | ||
+ | Now that we know <math>x</math> 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 <math>-\frac{8\sqrt3}{3}</math>. | ||
+ | Now, divide ABCDEF into two congruent triangles and a parallelogram: AFE, BCD, and ABDE. The areas of AFE and BCD are each <math>\frac12\cdot\frac{8\sqrt3}{3}\cdot8=\frac{32\sqrt3}{3}</math>. The area of ABDE is <math>\frac12\cdot8\cdot\frac{10\sqrt3}{3}=\frac{80\sqrt3}3</math>. | ||
+ | The total area of the hexagon is <math>2\cdot\frac{32\sqrt3}3+\frac{80\sqrt3}3=\frac{144\sqrt3}{3}=48\sqrt3\rightarrow48+3=\boxed{051}</math> | ||
== See also == | == See also == |
Revision as of 18:49, 2 September 2017
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 .
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
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.