Difference between revisions of "2012 AIME II Problems/Problem 13"
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<cmath>11+11+655 = \framebox{677}.</cmath> | <cmath>11+11+655 = \framebox{677}.</cmath> | ||
+ | == Solution 5 == | ||
+ | |||
+ | [[Image:Screen Shot 2020-02-17 at 2.24.50 PM.png|500px]] | ||
+ | |||
+ | We create a diagram. Each of the red quadrilaterals are actually 60 degree rotations about point A. We easily get that <math>\overline{C E_1} = \sqrt{11}</math>, and <math> \overline{C E_3} = \sqrt{11}</math>. If we set <math>\angle B A D_2 = \theta</math>, we can start angle chasing. In particular, we will like to find <math>\angle D E_4 C</math>, and <math>\angle D E_2 C</math>, since then we will be able to set up some Law Of Cosines. <math>\angle D E_4 C = \angle D E_4 A + \angle A E_4 C = 90 - \frac{\theta}{2} + 30 + \frac{\theta}{2} = 120^{\circ}</math> That was convenient! We can do it with the other angle as well. <math>\angle D E_2 C = \angle D E_2 A - \angle C E_2 A = 90 - \frac{\theta}{2} - (30 - \frac{\theta}{2}) = 60^{\circ}</math>. That means we are able to set up Law of Cosines, on triangles <math>\triangle D E_4 C</math> and <math>\triangle D E_2 C</math>, with some really convenient angles. | ||
+ | Let <math>CE_2 = x</math>, and <math>CE_4 = y</math>. | ||
+ | <cmath>333 = 11 + x^2 - \sqrt{11} x</cmath> | ||
+ | <cmath>333 = 11 + y^2 + \sqrt{11} y</cmath> | ||
+ | We subtract and get: | ||
+ | <cmath>0 = (x+y)(x-y-\sqrt{11})</cmath> | ||
+ | x+y obviously can't be 0, so <math>x-y = \sqrt{11}</math> | ||
+ | We add and get: | ||
+ | <cmath>666 = 22 + x^2 + y^2 + \sqrt{11} (y-x)</cmath>. | ||
+ | <math>y-x = -\sqrt{11}</math>. Thus, we can fill in and solve. | ||
+ | <cmath>666 = 22 + x^2 + y^2 - 11</cmath> | ||
+ | <cmath>655 = x^2 + y^2</cmath> | ||
+ | Thus our answer is <math>C E_1^2 + C E_2^2 + C E_2^2 + C E_4^2 = 11 + 11 + C E_2^2 + C E_4^2 = 11 + 11 + x^2 + y^2 = 11 + 11 + 655 = \boxed{677}</math>. | ||
+ | |||
+ | -Alexlikemath | ||
== 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 19:30, 17 February 2020
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 (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 every point is 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
Solution 5
We create a diagram. Each of the red quadrilaterals are actually 60 degree rotations about point A. We easily get that , and . If we set , we can start angle chasing. In particular, we will like to find , and , since then we will be able to set up some Law Of Cosines. That was convenient! We can do it with the other angle as well. . That means we are able to set up Law of Cosines, on triangles and , with some really convenient angles. Let , and . We subtract and get: x+y obviously can't be 0, so We add and get: . . Thus, we can fill in and solve. Thus our answer is .
-Alexlikemath
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.