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Difference between revisions of "2012 AIME II Problems/Problem 13"
(→‎Solution: Simplified the solution by removing the cosine fraction)
Line 10: Line 10:
  
 
<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 = \frac{222 - 11}{222} = \frac{211}{222}</cmath>
+
<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>.
Line 29: Line 29:
  
 
The solution is:
 
The solution is:
−
<cmath>(222 - 222\;cos\;(120 + \theta) + 222 - 222\;cos\;\theta + 222 - 222\;cos\;(120 - \theta) + 222 - 222\;cos\;\theta</cmath>
+
<cmath>(E_1C)^2 + (E_3C)^2 + (E_2C)^2 + (E_4C)^2</cmath>
−
<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>
+
<cmath>= 222\;(1 - cos\;(120 + \theta)) + 222\;(1 - cos\;(120 - \theta)) + 222\;(1 - cos\;\theta) + 222\;(1 - cos\;\theta)</cmath>
−
<cmath>= 888 - 222\;(2\;cos\;120\;cos\;\theta - 2\;cos\;\theta)</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>= 888 - 222\;cos\;\theta</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>= 888 - 222 \cdot \frac{211}{222} = \framebox{677}.</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 21:00, 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,

\[\sqrt{11}^2 = \sqrt{111}^2 + \sqrt{111}^2 - 2 \cdot \sqrt{111} \cdot \sqrt{111} \cdot cos\;\theta\] \[11 = 222\;(1 - cos\;\theta)\]

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: \[(E_1C)^2 + (E_3C)^2 + (E_2C)^2 + (E_4C)^2\] \[= 222\;(1 - cos\;(120 + \theta)) + 222\;(1 - cos\;(120 - \theta)) + 222\;(1 - cos\;\theta) + 222\;(1 - cos\;\theta)\] \[= 222\;((1 - (cos\;120\;cos\;\theta - sin\;120\;sin\;\theta)) + (1 - (cos\;120\;cos\;\theta + sin\;120\;sin\;\theta)) + 2\;(1 -\;cos\;\theta))\] \[= 222\;(1 - cos\;120\;cos\;\theta + sin\;120\;sin\;\theta + 1 - cos\;120\;cos\;\theta - sin\;120\;sin\;\theta + 2 - 2\;cos\;\theta)\] \[= 222\;(1 + \frac{1}{2}\;cos\;\theta + 1 + \frac{1}{2}\;cos\;\theta + 2 - 2\;cos\;\theta)\] \[= 222\;(4 - cos\;\theta)\] \[= 666 + 222\;(1 - cos\;\theta)\] Substituting $11$ for $222\;(1 - cos\;\theta)$ gives the solution $666 + 11 = \framebox{677}.$

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
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