Difference between revisions of "2012 AIME I Problems/Problem 12"
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== Solution == | == Solution == | ||
+ | === Solution 1 === | ||
Without loss of generality, set <math>CB = 1</math>. Then, by the Angle Bisector Theorem on triangle <math>DCB</math>, we have <math>CD = \frac{8}{15}</math>. We apply the Law of Cosines to triangle <math>DCB</math> to get <math>1 + \frac{64}{225} - \frac{8}{15} = BD^{2}</math>, which we can simplify to get <math>BD = \frac{13}{15}</math>. | Without loss of generality, set <math>CB = 1</math>. Then, by the Angle Bisector Theorem on triangle <math>DCB</math>, we have <math>CD = \frac{8}{15}</math>. We apply the Law of Cosines to triangle <math>DCB</math> to get <math>1 + \frac{64}{225} - \frac{8}{15} = BD^{2}</math>, which we can simplify to get <math>BD = \frac{13}{15}</math>. | ||
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Our final answer is <math>4+3+11 = \boxed{018}</math>. | Our final answer is <math>4+3+11 = \boxed{018}</math>. | ||
+ | |||
+ | === Solution 2 === | ||
+ | |||
+ | (This solution does not use the Angle Bisector Theorem but uses more trig.) | ||
+ | |||
+ | Find values for all angles in terms of <math>\angle B</math>. <math>\angle CEB = 150-B</math>, <math>\angle CED = 30+B</math>, <math>\angle CDE = 120-B</math>, <math>\angle CDA = 60+B</math>, and <math>\angle A = 90-B</math>. | ||
+ | |||
+ | Use the law of sines on <math>\triangle CED</math> and <math>\triangle CEB</math>: | ||
+ | |||
+ | In <math>\triangle CED</math>, <math>\frac{8}{\sin 30} = \frac{CE}{\sin (120-B)}</math>. This simplifies to <math>16 = \frac{CE}{\sin (120-B)}</math>. | ||
+ | |||
+ | In <math>\triangle CEB</math>, <math>\frac{15}{\sin 30} = \frac{CE}{\sin B}</math>. This simplifies to <math>30 = \frac{CE}{\sin B}</math>. | ||
+ | |||
+ | Solve for <math>CE</math> and equate them so that you get <math>16\sin (120-B) = 30\sin B</math>. | ||
+ | |||
+ | From this, <math>\frac{8}{15} = \frac{\sin B}{\sin (120-B)}</math>. | ||
+ | |||
+ | Use a trig identity on the denominator on the right to obtain: <math>\frac{8}{15} = \frac{\sin B}{\sin 120 \cos B - \cos 120 \sin B}</math> | ||
+ | |||
+ | This simplifies to <math>\frac{8}{15} = \frac{\sin B}{\frac{\sqrt{3}\cos B}{2} + \frac{\sin B}{2}} = \frac{\sin B}{\frac{\sqrt{3} \cosB + \sin B}{2}} = \frac{2\sin B}{\sqrt{3}\cos B + \sin B}</math> | ||
+ | |||
+ | This gives <math>\frac{15}{8} = \frac{\sqrt{3}\cos B + \sin B}{2\sin B} = \frac{\sqrt{3}\cos B}{2\sinB} + \frac{1}{2} = \frac{\sqrt{3}\cot B}{2} + \frac{1}{2}.</math> | ||
+ | |||
+ | Thus: <math>\frac{11}{7} = \frac{\sqrt{3}\cot B}{2}</math> and <math>\cot B = \frac{11*2}{8\sqrt{3}}</math>. | ||
+ | |||
+ | Since <math>\cot B = \frac{1}{\tan B}</math>, <math>tan{B} = \frac{8\sqrt{3}}{22} = \frac{4\sqrt{3}}{11}</math>. Our final answer is <math>4 + 3 + 11 = 18</math>. | ||
== See also == | == See also == | ||
{{AIME box|year=2012|n=I|num-b=11|num-a=13}} | {{AIME box|year=2012|n=I|num-b=11|num-a=13}} |
Revision as of 11:14, 18 March 2012
Problem 12
Let be a right triangle with right angle at Let and be points on with between and such that and trisect If then can be written as where and are relatively prime positive integers, and is a positive integer not divisible by the square of any prime. Find
Solution
Solution 1
Without loss of generality, set . Then, by the Angle Bisector Theorem on triangle , we have . We apply the Law of Cosines to triangle to get , which we can simplify to get .
Now, we have by another application of the Law of Cosines to triangle , so . In addition, , so .
Our final answer is .
Solution 2
(This solution does not use the Angle Bisector Theorem but uses more trig.)
Find values for all angles in terms of . , , , , and .
Use the law of sines on and :
In , . This simplifies to .
In , . This simplifies to .
Solve for and equate them so that you get .
From this, .
Use a trig identity on the denominator on the right to obtain:
This simplifies to $\frac{8}{15} = \frac{\sin B}{\frac{\sqrt{3}\cos B}{2} + \frac{\sin B}{2}} = \frac{\sin B}{\frac{\sqrt{3} \cosB + \sin B}{2}} = \frac{2\sin B}{\sqrt{3}\cos B + \sin B}$ (Error compiling LaTeX. ! Undefined control sequence.)
This gives $\frac{15}{8} = \frac{\sqrt{3}\cos B + \sin B}{2\sin B} = \frac{\sqrt{3}\cos B}{2\sinB} + \frac{1}{2} = \frac{\sqrt{3}\cot B}{2} + \frac{1}{2}.$ (Error compiling LaTeX. ! Undefined control sequence.)
Thus: and .
Since , . Our final answer is .
See also
2012 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 11 |
Followed by Problem 13 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |