Difference between revisions of "2012 AIME II Problems/Problem 13"
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Angle <math>E_1AC = E_1AD_1 + D_1AB + BAC = 60 + \theta + 60 = 120 + \theta</math>; by the Law of Cosines, | Angle <math>E_1AC = E_1AD_1 + D_1AB + BAC = 60 + \theta + 60 = 120 + \theta</math>; by the Law of Cosines, | ||
− | <cmath>\begin{align*}(E_1C)^2 &= (E_1A)^2 + (AC)^2 - 2 (E_1A) ( | + | <cmath>\begin{align*}(E_1C)^2 &= (E_1A)^2 + (AC)^2 - 2 (E_1A) (AC)\cos(120 + \theta)\\ |
&= 111 + 111 - 222\cos(120 + \theta)\end{align*}</cmath> | &= 111 + 111 - 222\cos(120 + \theta)\end{align*}</cmath> | ||
Angle <math>E_2AC = \theta</math>; by the Law of Cosines, | Angle <math>E_2AC = \theta</math>; by the Law of Cosines, | ||
− | <cmath>\begin{align*}(E_2C)^2 &= (E_2A)^2 + (AC)^2 - 2 (E_2A) ( | + | <cmath>\begin{align*}(E_2C)^2 &= (E_2A)^2 + (AC)^2 - 2 (E_2A) (AC)\cos\theta\\ |
&= 111 + 111 - 222\,\cos\theta\end{align*}</cmath> | &= 111 + 111 - 222\,\cos\theta\end{align*}</cmath> | ||
Line 22: | Line 22: | ||
Angle <math>E_3AC = E_3AB + BAC = (60 - \theta) + 60 = 120 - \theta</math>; by the Law of Cosines, | Angle <math>E_3AC = E_3AB + BAC = (60 - \theta) + 60 = 120 - \theta</math>; by the Law of Cosines, | ||
− | <cmath>\begin{align*}(E_3C)^2 &= (E_3A)^2 + (AC)^2 - 2 (E_3A) ( | + | <cmath>\begin{align*}(E_3C)^2 &= (E_3A)^2 + (AC)^2 - 2 (E_3A) (AC)\cos(120 - \theta)\\ |
&= 111 + 111 - 222\cos(120 - \theta)\end{align*}</cmath> | &= 111 + 111 - 222\cos(120 - \theta)\end{align*}</cmath> | ||
Angle <math>E_4AC = \theta</math>; by the Law of Cosines, | Angle <math>E_4AC = \theta</math>; by the Law of Cosines, | ||
− | <cmath>\begin{align*}(E_4C)^2 &= (E_4A)^2 + (AC)^2 - 2 (E_4A) ( | + | <cmath>\begin{align*}(E_4C)^2 &= (E_4A)^2 + (AC)^2 - 2 (E_4A) (AC)\cos\theta \\ |
&= 111 + 111 - 222\cos\theta\end{align*}</cmath> | &= 111 + 111 - 222\cos\theta\end{align*}</cmath> | ||
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-Alexlikemath | -Alexlikemath | ||
+ | |||
+ | ==Video Solution by MOP 2024== | ||
+ | https://youtu.be/IUOqCEGgG0g | ||
+ | |||
+ | ~r00tsOfUnity | ||
== See Also == | == See Also == |
Latest revision as of 18:02, 5 February 2024
Contents
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 primative 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:
obviously can't be 0, so
We add and get:
.
. Thus, we can fill in and solve.
Thus our answer is
.
-Alexlikemath
Video Solution by MOP 2024
~r00tsOfUnity
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.