Difference between revisions of "2012 AIME I Problems/Problem 12"
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=== Solution 2 === | === Solution 2 === | ||
− | (This solution does not use the Angle Bisector Theorem or the Law of Cosines, but it uses the Law of Sines and more trig | + | (This solution does not use the Angle Bisector Theorem or the Law of Cosines, but it uses the Law of Sines and 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>. | 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>. | ||
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Thus: <math>\frac{11}{7} = \frac{\sqrt{3}\cot B}{2}</math> and <math>\cot B = \frac{11*2}{8\sqrt{3}}</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 = | + | 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 = \boxed{018}</math>. |
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+ | === Solution 3 === | ||
+ | (This solution avoids advanced trigonometry) | ||
+ | |||
+ | Let <math>X</math> be the foot of the perpendicular from <math>D</math> to <math>\overline{BC}</math>, and let <math>Y</math> be the foot of the perpendicular from <math>E</math> to <math>\overline{BC}</math>. | ||
+ | |||
+ | Now let <math>EY=x</math>. Clearly, triangles <math>EYB</math> and <math>DXB</math> are similar with <math>\frac{BE}{BD}=\frac{15}{15+8}=\frac{15}{23}=\frac{EY}{DX}</math>, so <math>DX=\frac{23}{15}x</math>. | ||
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+ | Since triangles <math>CDX</math> and <math>CEY</math> are 30-60-90 right triangles, we can easily find other lengths in terms of <math>x</math>. For example, we see that <math>CY=x\sqrt{3}</math> and <math>CX=\frac{\frac{23}{15}x}{\sqrt{3}}=\frac{23\sqrt{3}}{45}x</math>. Therefore <math>XY=CY-CX=x\sqrt{3}-\frac{23\sqrt{3}}{45}x=\frac{22\sqrt{3}}{45}x</math>. | ||
+ | |||
+ | Again using the fact that triangles <math>EYB</math> and <math>DXB</math> are similar, we see that <math>\frac{BX}{BY}=\frac{XY+BY}{BY}=\frac{XY}{BY}+1=\frac{23}{15}</math>, so <math>BY=\frac{15}{8}XY=\frac{15}{8}*\frac{22\sqrt{3}}{45}=\frac{11\sqrt{3}}{2}</math>. | ||
+ | |||
+ | Thus <math>\tan \angle B = \frac{x}{\frac{11\sqrt{3}}{12}x}=\frac{4\sqrt{3}}{11}</math>, and our answer is <math>4+3+11=\boxed{018}</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 20:10, 1 March 2013
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 or the Law of Cosines, but it uses the Law of Sines and 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 .
Solution 3
(This solution avoids advanced trigonometry)
Let be the foot of the perpendicular from to , and let be the foot of the perpendicular from to .
Now let . Clearly, triangles and are similar with , so .
Since triangles and are 30-60-90 right triangles, we can easily find other lengths in terms of . For example, we see that and . Therefore .
Again using the fact that triangles and are similar, we see that , so .
Thus , and our 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 |