Difference between revisions of "2013 AIME I Problems/Problem 12"
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Evaluating and reducing, we get <math>\frac{9 + 5\sqrt{3}}{4}, </math>thus the answer is <math> \boxed{021}</math> | Evaluating and reducing, we get <math>\frac{9 + 5\sqrt{3}}{4}, </math>thus the answer is <math> \boxed{021}</math> | ||
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+ | ==Solution 3(Trig with Diagram)== | ||
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+ | [[File:2013_AIME_I_Problem_12.png]] | ||
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+ | With some simple angle chasing we can show that <math>\triangle OJL</math> and <math>\triangle MPL</math> are congruent. This means we have a large equilateral triangle with side length <math>3</math> and quadrilateral <math>OJQN</math>. We know that <math>[OJQN] = [\triangle NQL] - [\triangle OJL]</math>. Using Law of Sines and the fact that <math>\angle N = 45^{\circ}</math> we know that <math>\overline{NL} = \sqrt{6}</math> and the height to that side is <math>\frac{\sqrt{3} -1}{\sqrt{2}}</math> so <math>[\triangle NQL] = \frac{3-\sqrt{3}}{2}</math>. Using an extremely similar process we can show that <math>\overline{OJ} = 2-\sqrt{3}</math> which means the height to <math>\overline{LJ}</math> is <math>\frac{2\sqrt{3}-3}{2}</math>. So the area of <math>\triangle OJL = \frac{2\sqrt{3}-3}{4}</math>. This means the area of quadrilateral <math>OJQN = \frac{3-\sqrt{3}}{2} - \frac{2\sqrt{3}-3}{4} = \frac{9-4\sqrt{3}}{4}</math>. So the area of our larger triangle is <math>\frac{9-4\sqrt{3}}{4} + \frac{9\sqrt{3}}{4} = \frac{9+5\sqrt{3}}{4}</math>. Therefore <math>9+5+3+4 = \boxed{021}</math> | ||
== See also == | == See also == | ||
{{AIME box|year=2013|n=I|num-b=11|num-a=13}} | {{AIME box|year=2013|n=I|num-b=11|num-a=13}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 00:59, 3 September 2018
Contents
Problem 12
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
.
Cartesian Variation Solution
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 ordinate 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 2 (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 3(Trig with Diagram)
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
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.