Difference between revisions of "2012 AIME I Problems/Problem 12"
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== Solutions == | == Solutions == | ||
=== Solution 1 === | === Solution 1 === | ||
− | Let <math>CD = 2a</math>. Using angle bisector theorem on <math>\triangle CDB</math>, <math>\frac{2a}{8}=\frac{CB}{15}</math>, so <math>CB = \frac{15a}{4}</math>. Then, drop the altitude from <math>D</math> to <math>CB</math> and call the foot <math>F</math>. Thus, <math>CF = a</math>, <math> | + | Let <math>CD = 2a</math>. Using angle bisector theorem on <math>\triangle CDB</math>, <math>\frac{2a}{8}=\frac{CB}{15}</math>, so <math>CB = \frac{15a}{4}</math>. Then, drop the altitude from <math>D</math> to <math>CB</math> and call the foot <math>F</math>. Thus, <math>CF = a</math>, <math>FD = a\sqrt{3}</math>, and <math>FB = \frac{11a}{4}</math>. Finally, <math>\tan{B} = \frac{a\sqrt{3}}{\frac{11a}{4}} = \frac{4\sqrt{3}}{11}</math>. Our answer is <math>\boxed{018}</math>. |
=== Solution 2 === | === Solution 2 === |
Revision as of 15:58, 26 February 2016
Contents
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
Solutions
Solution 1
Let . Using angle bisector theorem on , , so . Then, drop the altitude from to and call the foot . Thus, , , and . Finally, . Our answer is .
Solution 2
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 3
(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
This gives Dividing by , we have
. Our final answer is .
Solution 4
(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 .
Solution 5
(Another solution without trigonometry)
Extend to point such that . It is then clear that is similar to .
Let , . Then .
With the Angle Bisector Theorem, we get that . From 30-60-90 , we get that and .
From , we have that . Simplifying yields , and , so 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.