Difference between revisions of "2012 AIME II Problems/Problem 13"
m (→Solution 2) |
(added solution 4) |
||
Line 95: | Line 95: | ||
<cmath>22+444+2\sqrt{111}Re(x) = 22+444+211 = \framebox{677}.</cmath> | <cmath>22+444+2\sqrt{111}Re(x) = 22+444+211 = \framebox{677}.</cmath> | ||
+ | |||
+ | ==Solution 4== | ||
+ | This method uses the observation that all the points are equidistant from <math>A</math>. Without loss of generality, we can assume <math>C</math> is on the same side of <math>AB</math> as <math>D_1</math>. | ||
+ | |||
+ | We can start off by angle chasing the angles around <math>A</math>. We let <math>\angle BAD_1 = \alpha</math>. Then, we note that <math>BD_1 = BD_2</math> and <math>AD_1 = AB = AD_2 </math>. Thus, <math>\bigtriangleup ABD_1 \cong \bigtriangleup ABD_2</math>. Thus, <math>\angle BAD_2 = \alpha</math> also. | ||
+ | |||
+ | We can now angle chase the angles about <math>A</math>. Because <math>\angle D_2AE_3 = 60</math>, <math>\angle D_1AE_3 = 60- 2 \alpha</math>. We can use all the congruent equilateral triangles in a similar manner obtaining: | ||
+ | <cmath>\angle D_1AE_3 = 60- 2 \alpha</cmath> | ||
+ | <cmath>\angle E_3AC = \alpha</cmath> | ||
+ | <cmath>\angle CAE_1 = \alpha</cmath> | ||
+ | <cmath>\angle D_2AE_2 = 60- 2 \alpha</cmath> | ||
+ | <cmath>\angle E_2AE_4 = 2 \alpha</cmath> | ||
+ | |||
+ | Now, <math>AE_3 = AC = \sqrt{111} </math> and <math>\angle E_3AC = \alpha </math>. Thus, <math>\bigtriangleup E_3AC \cong \bigtriangleup BAD_1</math>. Thus, <math>CE_3^2 = BD_1^2 = 11</math>. | ||
+ | |||
+ | Similarly, <math>AC = AE_1 = \sqrt{111} </math> and <math>\angle CAE_1 = \alpha </math>. Thus, <math>\bigtriangleup CAE_1 \cong \bigtriangleup BAD_1</math>. Thus, <math>CE_1^2 = BD_1^2 = 11</math>. | ||
+ | |||
+ | We can use <math>\bigtriangleup CAE_2</math> to find <math>CE_2^2</math>. Law of Cosines yields | ||
+ | <cmath> CE_2^2 = AE_2^2 + AC^2 - 2 \cdot AE_2 \cdot AC \cdot cos(\angle E_2AC).</cmath> | ||
+ | Substituting the known lengths and angles gives | ||
+ | <cmath> CE_2^2 = 222 - 222 \cdot cos(120- \alpha).</cmath> | ||
+ | Expanding this with the Cosine Subtraction Identity we get | ||
+ | <cmath> CE_2^2 = 222 - 222(cos120 cos \alpha + sin120 sin \alpha).</cmath> | ||
+ | |||
+ | We could attempt to calculate this but we can clear it up by simultaneously finding <math>CE_4^2</math> too. | ||
+ | We use Law of Cosines on <math>\bigtriangleup CAE_4</math> to get | ||
+ | <cmath> CE_4^2 = AE_4^2 + AC^2 - 2 \cdot AE_4 \cdot AC \cdot cos(\angle E_4AC).</cmath> | ||
+ | Substituting the known lengths and angles gives | ||
+ | <cmath> CE_4^2 = 222 - 222 \cdot cos(120 + \alpha).</cmath> | ||
+ | Expanding this with the Cosine Addition Identity we get | ||
+ | <cmath> CE_4^2 = 222 - 222(cos120 cos \alpha - sin120 sin \alpha).</cmath> | ||
+ | Adding this to our equation for <math>CE_2^2</math>, we get | ||
+ | <cmath> CE_2^2 + CE_4^2 = 444 - 222(cos120 cos \alpha - sin120 sin \alpha) - 222(cos120 cos \alpha + sin120 sin \alpha).</cmath> | ||
+ | Simplifying we get | ||
+ | <cmath> CE_2^2 + CE_4^2 = 444 - 444(cos120 cos \alpha)</cmath> | ||
+ | |||
+ | We can find <math>cos \alpha</math> by using Law of Cosines on <math>\bigtriangleup BAD_1</math>. This gives | ||
+ | <cmath> 11 = 222 - 222cos \alpha.</cmath> | ||
+ | Thus <math>cos \alpha = \frac{211}{222}</math>. | ||
+ | Substituting it in gives | ||
+ | <cmath> CE_2^2 + CE_4^2 = 444 - 444(cos120 \cdot \frac{211}{222}).</cmath> | ||
+ | Thus | ||
+ | <cmath> CE_2^2 + CE_4^2 = 444 + 211 = 655.</cmath> | ||
+ | |||
+ | Therefore the desired sum is equal to | ||
+ | |||
+ | <cmath>11+11+655 = \framebox{677}.</cmath> | ||
== 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}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 12:16, 14 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:
,
,
,
.
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
Solution 4
This method uses the observation that all the points are equidistant from . Without loss of generality, we can assume is on the same side of as .
We can start off by angle chasing the angles around . We let . Then, we note that and . Thus, . Thus, also.
We can now angle chase the angles about . Because , . We can use all the congruent equilateral triangles in a similar manner obtaining:
Now, and . Thus, . Thus, .
Similarly, and . Thus, . Thus, .
We can use to find . Law of Cosines yields Substituting the known lengths and angles gives Expanding this with the Cosine Subtraction Identity we get
We could attempt to calculate this but we can clear it up by simultaneously finding too. We use Law of Cosines on to get Substituting the known lengths and angles gives Expanding this with the Cosine Addition Identity we get Adding this to our equation for , we get Simplifying we get
We can find by using Law of Cosines on . This gives Thus . Substituting it in gives Thus
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.