Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution: Simplified the solution by removing the cosine fraction)
(→‎Solution)
Line 37: Line 37:
 
<cmath>= 666 + 222\;(1 - 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>
 
Substituting <math>11</math> for <math>222\;(1 - cos\;\theta)</math> gives the solution <math>666 + 11 = \framebox{677}.</math>
 +
 +
== Solution 2 ==
 +
This problem is pretty much destroyed by complex plane geometry, which is similar to vector geometry only with the power of easy rotation.  Place the triangle in the complex plane by letting <math>C</math> be the origin, placing <math>B</math> along the x-axis, and <math>A</math> in the first quadrant.  Let <math>r=\sqrt{111}</math>.  If <math>\omega</math> denotes the sixth root of unity, <math>e^{i\pi/3}</math>, then we have <math>C=0</math>, <math>B=r</math>, and <math>A=r\omega.</math>  Recall that counter-clockwise rotation in the complex plane by an angle <math>\theta</math> is accomplished by multiplication by <math>e^{i\theta}</math> (and clockwise rotation is multiplication by its conjugate).  So, we can find <math>D_1</math> and <math>D_2</math> by rotating <math>B</math> around <math>A</math> by angles of <math>\theta</math> and <math>-\theta</math>, where <math>\theta</math> is the apex angle in the isoceles triangle with sides <math>\sqrt{111}</math>, <math>\sqrt{111}</math>, and <math>\sqrt{11}</math>.  That is, let <math>z=e^{i\theta}</math>, and then:
 +
 +
<math>D_1=A+z(B-A)</math>, and <math>D_2=A+\overline{z}(B-A)</math>.  Now notice that <math>B-A=\overline{A}</math>, so this simplifies further to:
 +
 +
<math>D_1=A+z\overline{A}</math>, and <math>D_2=A+\overline{z}\overline{A}</math>. 
 +
 +
Similarly, we can write <math>E_1</math>, <math>E_2</math>, <math>E_3</math>, and <math>E_4</math> by rotating <math>D_1</math> and <math>D_2</math> around <math>A</math> by <math>\pm\pi/3</math>:
 +
 +
<math>E_1=A+\omega(D_1-A)</math>, <math>E_2=A+\overline{\omega}(D_1-A)</math>, <math>E_3=A+\omega(D_2-A)</math>, <math>E_4=A+\overline{\omega}(D_2-A)</math>.  Thus:
 +
 +
<math>E_1=A+\omega z \overline{A}</math>, <math>E_2=A+\overline{\omega} z \overline{A}</math>, <math>E_3=A+\omega\overline{z}\overline{A}</math>, <math>E_4=A+\overline{\omega}\overline{z}\overline{A}</math>.
 +
 +
Now to find some magnitudes, which is easy since we chose <math>C</math> as the origin:
 +
 +
<math>\|E_1\|^2=(A+\omega z \overline{A})(\overline{A}+\overline{\omega z}A)=2\|A\|^2+\omega z \overline{A}^2 + \overline{\omega}\overline{z}A^2</math>,
 +
 +
<math>\|E_2\|^2=(A+\overline{\omega} z \overline{A})(\overline{A}+\omega} \overline{z}A)=2\|A\|^2+\overline{\omega} z \overline{A}^2 + \omega\overline{z}A^2</math>,
 +
 +
<math>\|E_3\|^2=(A+\omega \overline{z} \overline{A})(\overline{A}+\overline{\omega} z A)=2\|A\|^2+\omega \overline{z} \overline{A}^2 + \overline{\omega}zA^2</math>,
 +
 +
<math>\|E_4\|^2=(A+\overline{\omega z} \overline{A})(\overline{A}+\omega zA)=2\|A\|^2+\overline{\omega z} \overline{A}^2 + \omega zA^2</math>.
 +
 +
Adding these up, the sum equals <math>8\|A\|^2+(\overline{A}^2+A^2)(\omega z + \omega \overline{z} + \overline{\omega}z + \overline{\omega}\overline{z}) = 8\|A\|^2+(\overline{A}^2+A^2)(z + \overline{z})( \omega+ \overline{\omega}) </math>. 
 +
 +
(Isn't that nice?)  We enter the home stretch by noticing that <math>\overline{A}^2+A^2 = r^2(\overline{\omega}^2+\omega^2) = -r^2</math>, and <math>\omega+ \overline{\omega}=1</math>.  Finally, <math>z + \overline{z} = 2\cos{\theta}</math>, which we can find using the law of cosines on that isoceles triangle: <math>11=111+111-222\cos{\theta}</math>, so <math>2\cos{\theta=211/111</math>.
 +
 +
Thus, the sum equals <math>888-211=\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 10:42, 19 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 $666 + 11 = \framebox{677}.$

Solution 2

This problem is pretty much destroyed by complex plane geometry, which is similar to vector geometry only with the power of easy rotation. Place the triangle in the complex plane by letting $C$ be the origin, placing $B$ along the x-axis, and $A$ in the first quadrant. Let $r=\sqrt{111}$. If $\omega$ denotes the sixth root of unity, $e^{i\pi/3}$, then we have $C=0$, $B=r$, and $A=r\omega.$ Recall that counter-clockwise rotation in the complex plane by an angle $\theta$ is accomplished by multiplication by $e^{i\theta}$ (and clockwise rotation is multiplication by its conjugate). So, we can find $D_1$ and $D_2$ by rotating $B$ around $A$ by angles of $\theta$ and $-\theta$, where $\theta$ is the apex angle in the isoceles triangle with sides $\sqrt{111}$, $\sqrt{111}$, and $\sqrt{11}$. That is, let $z=e^{i\theta}$, and then:

$D_1=A+z(B-A)$, and $D_2=A+\overline{z}(B-A)$. Now notice that $B-A=\overline{A}$, so this simplifies further to:

$D_1=A+z\overline{A}$, and $D_2=A+\overline{z}\overline{A}$.

Similarly, we can write $E_1$, $E_2$, $E_3$, and $E_4$ by rotating $D_1$ and $D_2$ around $A$ by $\pm\pi/3$:

$E_1=A+\omega(D_1-A)$, $E_2=A+\overline{\omega}(D_1-A)$, $E_3=A+\omega(D_2-A)$, $E_4=A+\overline{\omega}(D_2-A)$. Thus:

$E_1=A+\omega z \overline{A}$, $E_2=A+\overline{\omega} z \overline{A}$, $E_3=A+\omega\overline{z}\overline{A}$, $E_4=A+\overline{\omega}\overline{z}\overline{A}$.

Now to find some magnitudes, which is easy since we chose $C$ as the origin:

$\|E_1\|^2=(A+\omega z \overline{A})(\overline{A}+\overline{\omega z}A)=2\|A\|^2+\omega z \overline{A}^2 + \overline{\omega}\overline{z}A^2$,

$\|E_2\|^2=(A+\overline{\omega} z \overline{A})(\overline{A}+\omega} \overline{z}A)=2\|A\|^2+\overline{\omega} z \overline{A}^2 + \omega\overline{z}A^2$ (Error compiling LaTeX. Unknown error_msg),

$\|E_3\|^2=(A+\omega \overline{z} \overline{A})(\overline{A}+\overline{\omega} z A)=2\|A\|^2+\omega \overline{z} \overline{A}^2 + \overline{\omega}zA^2$,

$\|E_4\|^2=(A+\overline{\omega z} \overline{A})(\overline{A}+\omega zA)=2\|A\|^2+\overline{\omega z} \overline{A}^2 + \omega zA^2$.

Adding these up, the sum equals $8\|A\|^2+(\overline{A}^2+A^2)(\omega z + \omega \overline{z} + \overline{\omega}z + \overline{\omega}\overline{z}) = 8\|A\|^2+(\overline{A}^2+A^2)(z + \overline{z})( \omega+ \overline{\omega})$.

(Isn't that nice?) We enter the home stretch by noticing that $\overline{A}^2+A^2 = r^2(\overline{\omega}^2+\omega^2) = -r^2$, and $\omega+ \overline{\omega}=1$. Finally, $z + \overline{z} = 2\cos{\theta}$, which we can find using the law of cosines on that isoceles triangle: $11=111+111-222\cos{\theta}$, so $2\cos{\theta=211/111$ (Error compiling LaTeX. Unknown error_msg).

Thus, the sum equals $888-211=\framebox{677}$.

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
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2012_AIME_II_Problems/Problem_13&oldid=46469"