Difference between revisions of "2012 AIME II Problems/Problem 13"

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== Solution ==
 
== Solution ==
{{solution}}
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Note that there are only two possible locations for points <math>D_1</math> and <math>D_2</math>, as they are both <math>\sqrt{111}</math> from point <math>A</math> and <math>\sqrt{11}</math> from point <math>B</math>, so they are the two points where a circle centered at <math>A</math> with radius <math>\sqrt{111}</math> and a circle centered at <math>B</math> with radius <math>\sqrt{11}</math> intersect.  Let <math>D_1</math> be the point on the opposite side of <math>\overline{AB}</math> from <math>C</math>, and <math>D_2</math> the point on the same side of <math>\overline{AB}</math> as <math>C</math>.
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Let <math>\theta</math> be the measure of angle <math>BAD_1</math> (which is also the measure of angle <math>BAD_2</math>); by the Law of Cosines,
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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>
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<cmath>cos\;\theta = \frac{222 - 11}{222} = \frac{211}{222}</cmath>
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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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Angle <math>E_1AC = E_1AD_1 + D_1AB + BAC = 60 + \theta + 60 = 120 + \theta</math>; by the Law of Cosines,
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<cmath>(E_1C)^2 = (E_1A)^2 + (AC)^2 - 2 (E_1A) (E_1C)\;cos\;(120 + \theta)</cmath>
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<cmath>= 111 + 111 - 222\;cos\;(120 + \theta)</cmath>
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Angle <math>E_2AC = \theta</math>; by the Law of Cosines,
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<cmath>(E_2C)^2 = (E_2A)^2 + (AC)^2 - 2 (E_2A) (E_2C)\;cos\;\theta = 111 + 111 - 222\,cos\;\theta</cmath>
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There are two equilateral triangles with <math>\overline{AD_2}</math> as a side; let <math>E_3</math> be the third vertex that is farthest from <math>C</math>, and <math>E_4</math> be the third vertex that is nearest to <math>C</math>.
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Angle <math>E_3AC = E_3AB + BAC = (60 - \theta) + 60 = 120 - \theta</math>; by the Law of Cosines,
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<cmath>(E_3C)^2 = (E_3A)^2 + (AC)^2 - 2 (E_3A) (E_3C)\;cos\;(120 - \theta)</cmath>
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<cmath>= 111 + 111 - 222\;cos\;(120 - \theta)</cmath>
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Angle <math>E_4AC = \theta</math>; by the Law of Cosines,
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<cmath>(E_4C)^2 = (E_4A)^2 + (AC)^2 - 2 (E_4A) (E_4C)\;cos\;\theta = 111 + 111 - 222\;cos\;\theta</cmath>
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The solution is:
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<cmath>(222 - 222\;cos\;(120 + \theta) + 222 - 222\;cos\;\theta + 222 - 222\;cos\;(120 - \theta) + 222 - 222\;cos\;\theta</cmath>
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<cmath>= 888 - 222\;(cos\;120\;cos\;\theta - sin\;120\;sin\;\theta) - 222\;cos\;120\;cos\;\theta - 222 (cos\;120\;cos\;\theta + sin\;120\;sin\;\theta) - 222\;cos\;\theta)</cmath>
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<cmath>= 888 - 222\;(2\;cos\;120\;cos\;\theta - 2\;cos\;\theta)</cmath>
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<cmath>= 888 - 222\;cos\;\theta</cmath>
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<cmath>= 888 - 222 \cdot \frac{211}{222} = \framebox{677}.</cmath>
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== 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 12:13, 4 April 2012

Problem 13

Equilateral $\triangle ABC$ has side length $\sqrt{111}$. There are four distinct triangles $AD_1E_1$, $AD_1E_2$, $AD_2E_3$, and $AD_2E_4$, each congruent to $\triangle ABC$, with $BD_1 = BD_2 = \sqrt{11}$. Find $\sum_{k=1}^4(CE_k)^2$.


Solution

Note that there are only two possible locations for points $D_1$ and $D_2$, as they are both $\sqrt{111}$ from point $A$ and $\sqrt{11}$ from point $B$, so they are the two points where a circle centered at $A$ with radius $\sqrt{111}$ and a circle centered at $B$ with radius $\sqrt{11}$ intersect. Let $D_1$ be the point on the opposite side of $\overline{AB}$ from $C$, and $D_2$ the point on the same side of $\overline{AB}$ as $C$.

Let $\theta$ be the measure of angle $BAD_1$ (which is also the measure of angle $BAD_2$); by the Law of Cosines,

\[\sqrt{11}^2 = \sqrt{111}^2 + \sqrt{111}^2 - 2 \cdot \sqrt{111} \cdot \sqrt{111} \cdot cos\;\theta\] \[cos\;\theta = \frac{222 - 11}{222} = \frac{211}{222}\]

There are two equilateral triangles with $\overline{AD_1}$ as a side; let $E_1$ be the third vertex that is farthest from $C$, and $E_2$ be the third vertex that is nearest to $C$.

Angle $E_1AC = E_1AD_1 + D_1AB + BAC = 60 + \theta + 60 = 120 + \theta$; by the Law of Cosines, \[(E_1C)^2 = (E_1A)^2 + (AC)^2 - 2 (E_1A) (E_1C)\;cos\;(120 + \theta)\] \[= 111 + 111 - 222\;cos\;(120 + \theta)\] Angle $E_2AC = \theta$; by the Law of Cosines, \[(E_2C)^2 = (E_2A)^2 + (AC)^2 - 2 (E_2A) (E_2C)\;cos\;\theta = 111 + 111 - 222\,cos\;\theta\]

There are two equilateral triangles with $\overline{AD_2}$ as a side; let $E_3$ be the third vertex that is farthest from $C$, and $E_4$ be the third vertex that is nearest to $C$.

Angle $E_3AC = E_3AB + BAC = (60 - \theta) + 60 = 120 - \theta$; by the Law of Cosines, \[(E_3C)^2 = (E_3A)^2 + (AC)^2 - 2 (E_3A) (E_3C)\;cos\;(120 - \theta)\] \[= 111 + 111 - 222\;cos\;(120 - \theta)\] Angle $E_4AC = \theta$; by the Law of Cosines, \[(E_4C)^2 = (E_4A)^2 + (AC)^2 - 2 (E_4A) (E_4C)\;cos\;\theta = 111 + 111 - 222\;cos\;\theta\]

The solution is: \[(222 - 222\;cos\;(120 + \theta) + 222 - 222\;cos\;\theta + 222 - 222\;cos\;(120 - \theta) + 222 - 222\;cos\;\theta\] \[= 888 - 222\;(cos\;120\;cos\;\theta - sin\;120\;sin\;\theta) - 222\;cos\;120\;cos\;\theta - 222 (cos\;120\;cos\;\theta + sin\;120\;sin\;\theta) - 222\;cos\;\theta)\] \[= 888 - 222\;(2\;cos\;120\;cos\;\theta - 2\;cos\;\theta)\] \[= 888 - 222\;cos\;\theta\] \[= 888 - 222 \cdot \frac{211}{222} = \framebox{677}.\]

See Also

2012 AIME II (ProblemsAnswer KeyResources)
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
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