2013 AIME I Problems/Problem 12
Contents
Problem
Let be a triangle with
and
. A regular hexagon
with side length 1 is drawn inside
so that side
lies on
, side
lies on
, and one of the remaining vertices lies on
. There are positive integers
and
such that the area of
can be expressed in the form
, where
and
are relatively prime, and c is not divisible by the square of any prime. Find
.
Solution 1
First, find that .
Draw
. Now draw
around
such that
is adjacent to
and
. The height of
is
, so the length of base
is
. Let the equation of
be
. Then, the equation of
is
. Solving the two equations gives
. The area of
is
.
Solution 2 (Cartesian Variation)
Use coordinates. Call the origin and
be on the x-axis. It is easy to see that
is the vertex on
. After labeling coordinates (noting additionally that
is an equilateral triangle), we see that the area is
times
times the coordinate of
. Draw a perpendicular of
, call it
, and note that
after using the trig functions for
degrees.
Now, get the lines for and
:
and
, whereupon we get the ordinate of
to be
, and the area is
, so our answer is
.
Solution 3 (Trig)
Angle chasing yields that both triangles and
are
-
-
triangles. First look at triangle
. Using Law of Sines, we find:
Simplifying, we find .
Since
, WLOG assume triangle
is equilateral, so
. So
.
Apply Law of Sines again,
Simplifying, we find .
.
Evaluating and reducing, we get thus the answer is
Solution 4 (Special Triangles)
As we can see, the angle of
can be split into a
angle and a
angle. This allows us to drop an altitude from point
for
which intersects
at point
and
at point
. The main idea of our solution is to obtain enough sides of
that we are able to directly figure out its area (specifically by figuring out side
and
.
We first begin by figuring out the length of . This can be easily done, since
is simply
(given in the problem) and
because
is an equilateral after some simple angle calculations. Now we need to find
. This is when we bring in some simple algebra.
PREPARATION:
(45-45-90 Right Triangle)
(30-60-90 Right Triangle)
SOLVING:
so
Finally,
Now, we can finally get the length of by adding up
, which is simply
To get and
, we first work bit by bit.
(30-60-90 Right Triangle)
(same 30-60-90 Right Triangle)
Since because of 45-45-90 right triangles,
too.
Now, we can finally calculate , and it is
.
Finally, the area of can be calculated by
, which is equal to
. So the final answer is
.
Solution 5 (Trig)
With some simple angle chasing we can show that and
are congruent. This means we have a large equilateral triangle with side length
and quadrilateral
. We know that
. Using Law of Sines and the fact that
we know that
and the height to that side is
so
. Using an extremely similar process we can show that
which means the height to
is
. So the area of
. This means the area of quadrilateral
. So the area of our larger triangle is
. Therefore
.
Solution 6 (Elementary Geo)
We can find that . This means that the perpendicular from
to
is perpendicular to
as well, so let that perpendicular intersect
at
, and the perpendicular intersect
at
. Set
. Note that
, so
and
. Also,
, so
. It's easy to calculate the area now, because the perpendicular from
to
splits
into a
(PHQ) and a
(PHR). From these triangles' ratios, it should follow that
, so the area is
.
.
By Mathscienceclass
Solution 7 (Combination of 1 & 2)
We can observe that (because
&
are both
). Thus we know that
is equivalent to the height of the hexagon, which is
. Now we look at triangle
and apply the Law of Sines to it.
. From here we can solve for
and get that
. Now we use the Sine formula for the area of a triangle with sides
,
, and
to get the answer. Setting
and
we get the expression
which is
. Thus our final answer is
.
By AwesomeLife_Math
See also
2013 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 11 |
Followed by Problem 13 | |
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.