Difference between revisions of "2001 AMC 12 Problems/Problem 24"
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<math>\angle ADB = 120^\circ</math>, <math>\angle ADC = 60^\circ</math>, <math>\angle DAB = 15^\circ</math>, let <math>\angle ACB = \theta</math>, <math>\angle DAC = 120^\circ - \theta</math> | <math>\angle ADB = 120^\circ</math>, <math>\angle ADC = 60^\circ</math>, <math>\angle DAB = 15^\circ</math>, let <math>\angle ACB = \theta</math>, <math>\angle DAC = 120^\circ - \theta</math> | ||
− | By the [[Law of Sines]] we have <math>\frac{CD}{\sin(120^\circ - \theta)} = \frac{AD}{\sin \theta}</math> | + | By the [[Law of Sines]] we have <math>\frac{CD}{\sin(120^\circ - \theta)} = \frac{AD}{\sin \theta}</math> |
+ | |||
+ | <math>\space</math> <math>\frac{BD}{\sin15^\circ} = \frac{AD}{\sin45^\circ}</math> | ||
<math>\frac{BD}{CD} \cdot \frac{\sin(120^\circ - \theta)}{\sin15^\circ} = \frac{\sin \theta}{\sin45^\circ}</math> | <math>\frac{BD}{CD} \cdot \frac{\sin(120^\circ - \theta)}{\sin15^\circ} = \frac{\sin \theta}{\sin45^\circ}</math> | ||
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<math>\sin^2 15^\circ = \frac{1 - \cos 30^\circ}{2} = \frac{1 - \frac{\sqrt{3}}{2}}{2} = \frac{2 - \sqrt{3}}{4}</math> | <math>\sin^2 15^\circ = \frac{1 - \cos 30^\circ}{2} = \frac{1 - \frac{\sqrt{3}}{2}}{2} = \frac{2 - \sqrt{3}}{4}</math> | ||
− | <math>\frac{\sin(120^\circ - \theta)}{2} = \frac{\sin \theta}{\frac{2 - \sqrt{3}}{4}} = \frac{\sin \theta}{1+\sqrt{3}}</math> | + | <math>\frac{\sin(120^\circ - \theta)}{2} = \frac{\sin \theta}{\frac{2 - \sqrt{3}}{4}} = \frac{\sin \theta}{1+\sqrt{3}}</math>, <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{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> | ||
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<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> | <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 | + | 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> | Therefore <math>2 \theta = 150^\circ</math>, <math>\angle ACB = \boxed{\textbf{(D) } 75^\circ }</math> |
Revision as of 10:45, 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 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 |
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