Difference between revisions of "2012 AIME II Problems/Problem 13"
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(Isn't that nice?) Notice that <math>\overline{A}^2+A^2 = r^2(\overline{\omega}^2+\omega^2) = -r^2</math>, and <math>\omega+ \overline{\omega}=1</math>, so that this sum simplifies further to <math>888-111(z + \overline{z})</math>. | (Isn't that nice?) Notice that <math>\overline{A}^2+A^2 = r^2(\overline{\omega}^2+\omega^2) = -r^2</math>, and <math>\omega+ \overline{\omega}=1</math>, so that this sum simplifies further to <math>888-111(z + \overline{z})</math>. | ||
− | Finally, <math>z + \overline{z} = 2\cos{\theta}</math>, which is found 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>. | + | Finally, <math>z + \overline{z} = 2\cos{\theta}</math>, which is found 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>. | Thus, the sum equals <math>888-211=\framebox{677}</math>. |
Revision as of 21:35, 7 March 2015
Problem 13
Equilateral has side length . There are four distinct triangles , , , and , each congruent to , with . Find .
Solution 1
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
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 be the origin, placing along the x-axis, and in the first quadrant. Let . If denotes the sixth root of unity, , then we have , , and Recall that counter-clockwise rotation in the complex plane by an angle is accomplished by multiplication by (and clockwise rotation is multiplication by its conjugate). So, we can find and by rotating around by angles of and , where is the apex angle in the isoceles triangle with sides , , and . That is, let , and then:
, and . Now notice that , so this simplifies further to:
, and .
Similarly, we can write , , , and by rotating and around by :
, , , . Thus:
, , , .
Now to find some magnitudes, which is easy since we chose as the origin:
,
$\|E_2\|^2=(A+\overline{\omega} z \overline{A})(\overline{A}+\omega} \overline{z}A)=2\|A\|^2+\overline{\omega} z \overline{A}^2 + \omega\overline{z}A^2$ (Error compiling LaTeX. ! Extra }, or forgotten $.),
,
.
Adding these up, the sum equals .
(Isn't that nice?) Notice that , and , so that this sum simplifies further to .
Finally, , which is found using the law of cosines on that isoceles triangle: , so .
Thus, the sum equals .
Solution 3
This method uses complex numbers with as the origin. Let , , , where .
Also, let be or . Then
Therefore, , so
Since , are one of or , without loss of generality, let and . Then
One can similarly get and , so the desired sum is equal to
Note that , so the sum of these two is just . Therefore the desired sum is equal to
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.