Difference between revisions of "2013 AIME I Problems/Problem 12"
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Expilncalc (talk | contribs) (→Cartesian Variation Solution) |
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==Cartesian Variation Solution== | ==Cartesian Variation Solution== | ||
− | Also use a coordinate system, but upon drawing the diagram call <math>Q</math> the origin and <math>QP</math> be on the x-axis. | + | Also use a coordinate system, but upon drawing the diagram call <math>Q</math> the origin and <math>QP</math> be on the x-axis. Obviously <math>F</math> is the vertex on <math>RP</math>. After labeling coordinates (noting additionally that <math>QBC</math> is an equilateral triangle), we see that the area is <math>QP</math> times <math>0.5</math> times the ordinate of <math>R</math>. Draw a perpendicular of <math>F</math>, call it <math>H</math>, and note that <math>QP = 1 + \sqrt{3}</math> after using the trig functions for <math>75</math> degrees. |
Now, get the lines for <math>QR</math> and <math>RP</math>: <math>y=\sqrt{3}x</math> and <math>y=-(2+\sqrt{3})x + (5+\sqrt{3})</math>, whereupon we get the ordinate of <math>R</math> to be <math>\frac{3+2\sqrt{3}}{2}</math>, and the area is <math>\frac{5\sqrt{3} + 9}{4}</math>, so our answer is <math>\boxed{021}</math>. | Now, get the lines for <math>QR</math> and <math>RP</math>: <math>y=\sqrt{3}x</math> and <math>y=-(2+\sqrt{3})x + (5+\sqrt{3})</math>, whereupon we get the ordinate of <math>R</math> to be <math>\frac{3+2\sqrt{3}}{2}</math>, and the area is <math>\frac{5\sqrt{3} + 9}{4}</math>, so our 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 22:51, 23 January 2018
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
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
Also use a coordinate system, but upon drawing the diagram call the origin and
be on the x-axis. Obviously
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
.
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.