Difference between revisions of "2001 AMC 12 Problems/Problem 24"
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Finally, we get <math>\angle ACB = \angle ACE + \angle ECD = 45^\circ + 30^\circ = \boxed{75^\circ}</math>. | Finally, we get <math>\angle ACB = \angle ACE + \angle ECD = 45^\circ + 30^\circ = \boxed{75^\circ}</math>. | ||
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+ | ==Trig Bash== | ||
+ | WLOG, we can assume that <math>BD = 1</math> and <math>CD = 2</math>. As above, we are able to find that <math>\angle ADB = 60^\circ</math> and <math>\angle ADC = 120^\circ</math>. | ||
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+ | Using Law of Sines on triangle <math>ADB</math>, we find that <math>\frac{1}{\sin15^\circ} = \frac{AD}{\sin 45^\circ} = \frac{AB}{\sin 120^\circ}</math>. Since we know that <math>\sin 15^\circ = \frac{\sqrt{6}-\sqrt{2}}{4}</math>, <math>\sin 45^\circ = \frac{\sqrt{2}}{2}</math>, and <math>\sin 120^\circ = \frac{\sqrt{3}}{2}</math>, we can compute <math>AD</math> to equal <math>1+\sqrt{3}</math> and <math>AB</math> to be <math>\frac{3\sqrt{2}+\sqrt{6}}{2}</math>. | ||
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+ | Next, we apply Law of Cosines to triangle <math>ADC</math> to see that <math>AC^2 = (1+\sqrt{3})^2 + 2^2 - (2)(1+\sqrt{3})(2)(\cos 60^\circ)</math>. Simplifying the RHS, we get <math>AC^2 = 6</math>, so <math>AC = \sqrt{6}</math>. | ||
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+ | Now, we apply Law of Sines to triangle <math>ABC</math> to see that <math>\frac{\sqrt{6}}{\sin 45^\circ} = \frac{\frac{3\sqrt{2}+\sqrt{6}}{2}}{\sin ACB}</math>. After rearranging and noting that <math>\sin 45^\circ = \frac{\sqrt{2}}{2}</math>, we get <math>\sin ACB = \frac{\sqrt{6}+3\sqrt{2}}{4\sqrt{3}}</math>. | ||
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+ | Dividing the RHS through by <math>\sqrt{3}</math>, we see that <math>\sin ACB = \frac{\sqrt{6}+\sqrt{2}}{4}</math>, so <math>\angle ACB</math> is either <math>75^\circ</math> or <math>105^\circ</math>. Since <math>105^\circ</math> is not a choice, we know <math>\angle ACB = \boxed{75^\circ}</math>. | ||
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+ | Note that we can also confirm that <math>\angle ACB \neq 105^\circ</math> by computing <math>\angle CAB</math> with Law of Sines. | ||
== See Also == | == See Also == |
Revision as of 22:24, 1 November 2015
Contents
Problem
In , . Point is on so that and . Find .
Solution
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 .
Trig Bash
WLOG, 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 , , and , we can compute to equal and to be .
Next, we apply Law of Cosines to triangle to see that . Simplifying the RHS, we get , so .
Now, we apply Law of Sines to triangle to see that . After rearranging and noting that , we get .
Dividing the RHS through 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.
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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