Difference between revisions of "2001 AMC 12 Problems/Problem 24"
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<math>\frac{\sin \theta}{\sin(120^\circ - \theta)} = \frac{1+\sqrt{3}}{2}</math> | <math>\frac{\sin \theta}{\sin(120^\circ - \theta)} = \frac{1+\sqrt{3}}{2}</math> | ||
− | <math>\sin(120^\circ - \theta) = \sin120^\circ \cos\theta - \sin\theta \cos120^\circ = \frac{\sqrt{3}}{2} \cdot \cos\theta + \frac{\sqrt{1}}{2} \cdot \sin\theta</math> | + | <math>\sin(120^\circ - \theta) = \sin120^\circ \cos\theta - \sin\theta \cos120^\circ = \frac{\sqrt{3}}{2} \cdot \cos\theta + \frac{1}{2} \cdot \sin\theta</math> |
+ | |||
+ | <math>\frac{\sin \theta}{\sin(120^\circ - \theta)} = \frac{\sin \theta}{\frac{\sqrt{3}}{2} \cdot \cos\theta + \frac{\sqrt{1}}{2} \cdot \sin\theta} = \frac{1+\sqrt{3}}{2}</math> | ||
+ | |||
+ | <math>\frac{\sqrt{3} \cdot \cos\theta + \sin\theta}{2 \sin \theta} = \frac{2}{1+\sqrt{3}}</math> | ||
+ | |||
+ | <math>\frac{\sqrt{3}}{2} \cdot \cot \theta = \frac{2}{1+\sqrt{3}} - \frac{1}{2} = \frac{2 - \sqrt{3}}{2 + 2 \sqrt{3}}</math> | ||
+ | |||
+ | <math>\cot \theta = \frac{\cos \theta}{\sin \theta} = \frac{\sqrt{3} - 1}{\sqrt{3} + 1}</math> | ||
+ | |||
+ | Suppose <math>\cos \theta = k(\sqrt{3} - 1)</math>, and <math>\sin \theta = k(\sqrt{3} + 1)</math> | ||
+ | |||
+ | <math>\sin^2 \theta + \cos^2 \theta = 1</math>, <math>k^2(\sqrt{3} + 1)^2 + k^2(\sqrt{3} - 1)^2 = 8k^2 = 1</math>, <math>k = \frac{1}{2 \sqrt{2}}</math> | ||
+ | |||
+ | <math>\sin \theta = \frac{\sqrt{3} + 1}{2 \sqrt{2}}</math>, <math>\cos \theta = \frac{\sqrt{3} - 1}{2 \sqrt{2}}</math> | ||
+ | |||
+ | <math>\sin 2 \theta = 2 \cdot \frac{\sqrt{3} + 1}{2 \sqrt{2}} \cdot \frac{\sqrt{3} - 1}{2 \sqrt{2}} = \frac{1}{2}</math> | ||
+ | |||
+ | Two possible values of <math>2 \theta</math> are <math>150^\circ</math> and <math>30^\circ</math>. However we can rule out <math>30^\circ</math> because <math>\cos \15^\circ</math> is positive, while <math>\cos \theta</math> is negative. | ||
+ | |||
+ | Therefore <math>2 \theta = 150^\circ</math>, <math>\angle ACB = \boxed{\textbf{(D) } 75^\circ }</math> | ||
+ | |||
+ | ~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen] | ||
== See Also == | == See Also == |
Revision as of 11:42, 25 August 2022
Contents
Problem
In ,
. Point
is on
so that
and
. Find
Solution 1
We start with the observation that , and
.
We can draw the height from
onto
. In the triangle
, we have
. Hence
.
By the definition of , we also have
, therefore
. This means that the triangle
is isosceles, and as
, we must have
.
Then we compute , thus
and the triangle
is isosceles as well. Hence
.
Now we can note that , hence also the triangle
is isosceles and we have
.
Combining the previous two observations we get that , and as
, this means that
.
Finally, we get .
Solution 2
Draw a good diagram! Now, let's call , so
. Given the rather nice angles of
and
as you can see, let's do trig. Drop an altitude from
to
; call this point
. We realize that there is no specific factor of
we can call this just yet, so let
. Notice that in
we get
. Using the 60-degree angle in
, we obtain
. The comparable ratio is that
. If we involve our
, we get:
. Eliminating
and removing radicals from the denominator, we get
. From there, one can easily obtain
. Now we finally have a desired ratio. Since
upon calculation, we know that
can be simplified. Indeed, if you know that
or even take a minute or two to work out the sine and cosine using
, and perhaps the half- or double-angle formulas, you get
.
Solution 3
Without loss of generality, we can assume that and
. As above, we are able to find that
and
.
Using Law of Sines on triangle , we find that
Since we know that
we can compute
to equal
and
to be
.
Next, we apply Law of Cosines to triangle to see that
Simplifying the right side, we get
, so
.
Now, we apply Law of Sines to triangle to see that
After rearranging and noting that
, we get
Dividing the right side by , we see that
so
is either
or
. Since
is not a choice, we know
.
Note that we can also confirm that by computing
with Law of Sines.
Solution 4(FAST)
Note that and
. Seeing these angles makes us think of 30-60-90 triangles. Let
be the foot of the altitude from
to
. This means
and
. Let
and
. This means
and since
we know that
. This means
. This gives
. Note that
. Looking that the answer options we see that
. This means the answer is
.
~coolmath_2018
Solution 5 (Law of Sines)
,
,
, let
,
By the Law of Sines we have ,
By the Triple-angle identities,
Suppose , and
,
,
,
Two possible values of are
and
. However we can rule out
because $\cos \15^\circ$ (Error compiling LaTeX. Unknown error_msg) is positive, while
is negative.
Therefore ,
See Also
2001 AMC 12 (Problems • Answer Key • Resources) | |
Preceded by Problem 23 |
Followed by Problem 25 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.