Difference between revisions of "2012 AIME II Problems/Problem 13"
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<cmath>\sqrt{11}^2 = \sqrt{111}^2 + \sqrt{111}^2 - 2 \cdot \sqrt{111} \cdot \sqrt{111} \cdot cos\;\theta</cmath> | <cmath>\sqrt{11}^2 = \sqrt{111}^2 + \sqrt{111}^2 - 2 \cdot \sqrt{111} \cdot \sqrt{111} \cdot cos\;\theta</cmath> | ||
− | <cmath>cos\;\theta | + | <cmath>11 = 222\;(1 - cos\;\theta)</cmath> |
There are two equilateral triangles with <math>\overline{AD_1}</math> as a side; let <math>E_1</math> be the third vertex that is farthest from <math>C</math>, and <math>E_2</math> be the third vertex that is nearest to <math>C</math>. | There are two equilateral triangles with <math>\overline{AD_1}</math> as a side; let <math>E_1</math> be the third vertex that is farthest from <math>C</math>, and <math>E_2</math> be the third vertex that is nearest to <math>C</math>. | ||
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The solution is: | The solution is: | ||
− | <cmath>( | + | <cmath>(E_1C)^2 + (E_3C)^2 + (E_2C)^2 + (E_4C)^2</cmath> |
− | <cmath>= | + | <cmath>= 222\;(1 - cos\;(120 + \theta)) + 222\;(1 - cos\;(120 - \theta)) + 222\;(1 - cos\;\theta) + 222\;(1 - cos\;\theta)</cmath> |
− | <cmath>= | + | <cmath>= 222\;((1 - (cos\;120\;cos\;\theta - sin\;120\;sin\;\theta)) + (1 - (cos\;120\;cos\;\theta + sin\;120\;sin\;\theta)) + 2\;(1 -\;cos\;\theta))</cmath> |
− | <cmath>= | + | <cmath>= 222\;(1 - cos\;120\;cos\;\theta + sin\;120\;sin\;\theta + 1 - cos\;120\;cos\;\theta - sin\;120\;sin\;\theta + 2 - 2\;cos\;\theta)</cmath> |
− | <cmath>= | + | <cmath>= 222\;(1 + \frac{1}{2}\;cos\;\theta + 1 + \frac{1}{2}\;cos\;\theta + 2 - 2\;cos\;\theta)</cmath> |
+ | <cmath>= 222\;(4 - cos\;\theta)</cmath> | ||
+ | <cmath>= 666 + 222\;(1 - cos\;\theta)</cmath> | ||
+ | Substituting <math>11</math> for <math>222\;(1 - cos\;\theta)</math> gives the solution <math>666 + 11 = \framebox{677}.</math> | ||
== See Also == | == See Also == | ||
{{AIME box|year=2012|n=II|num-b=12|num-a=14}} | {{AIME box|year=2012|n=II|num-b=12|num-a=14}} |
Revision as of 22:00, 4 April 2012
Problem 13
Equilateral has side length
. There are four distinct triangles
,
,
, and
, each congruent to
,
with
. Find
.
Solution
Note that there are only two possible locations for points and
, as they are both
from point
and
from point
, so they are the two points where a circle centered at
with radius
and a circle centered at
with radius
intersect. Let
be the point on the opposite side of
from
, and
the point on the same side of
as
.
Let be the measure of angle
(which is also the measure of angle
); by the Law of Cosines,
There are two equilateral triangles with as a side; let
be the third vertex that is farthest from
, and
be the third vertex that is nearest to
.
Angle ; by the Law of Cosines,
Angle
; by the Law of Cosines,
There are two equilateral triangles with as a side; let
be the third vertex that is farthest from
, and
be the third vertex that is nearest to
.
Angle ; by the Law of Cosines,
Angle
; by the Law of Cosines,
The solution is:
Substituting
for
gives the solution
See Also
2012 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 12 |
Followed by Problem 14 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |