Difference between revisions of "2013 AIME I Problems/Problem 12"
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Simplifying, we find <math>PR = \frac{\sqrt{6}}{2} \cdot (1 + \sqrt{3})</math>. | Simplifying, we find <math>PR = \frac{\sqrt{6}}{2} \cdot (1 + \sqrt{3})</math>. | ||
− | <math>[PQR] = \frac{1}{2} \cdot PQ \cdot PR \cdot | + | <math>[PQR] = \frac{1}{2} \cdot PQ \cdot PR \cdot sin 75^\circ</math>. |
− | 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> |
== 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 18:52, 16 June 2018
Contents
[hide]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
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.