2005 AIME II Problems/Problem 14
Problem
In triangle and Point is on with Point is on such that Given that where and are relatively prime positive integers, find
Solution 1
By the Law of Sines and since , we have
Substituting our knowns, we have . The answer is .
Solution 2 (Similar Triangles)
Drop the altitude from A and call the base of the altitude Q. Also, drop the altitudes from E and D to AB and AC respectively. Call the feet of the altitudes R and S respectively.
From here, we can use Heron's Formula to find the altitude. The area of the triangle is = 84. We can then use similar triangles with triangle AQC and triangle DSC to find DS=. Consequently, from Pythagorean theorem, SC = and AS = 14-SC = . We can also use pythagorean triangle on triangle AQB yo determine that BQ = .
Label AR as y and RE as x. RB then equals 13-y. Then, we have two similar triangles.
Firstly: . From there, we have .
Next: . From there, we have .
Solve the system to get and . Notice that 463 is prime, so even though we use pythagorean theorem on x and 13-y, the denominator won't change. The answer we desire is .
See also
2005 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 13 |
Followed by Problem 15 | |
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