Difference between revisions of "2006 iTest Problems/Problem 32"
Rockmanex3 (talk | contribs) (Solution to Problem 32 -- Triangle Angle Trisection) |
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&= \frac{84\sqrt{10}}{5} | &= \frac{84\sqrt{10}}{5} | ||
\end{align*}</cmath> | \end{align*}</cmath> | ||
− | Thus, <math>\ | + | Thus, <math>\frac{AB}{AC} = \frac{84\sqrt{10}}{5} \cdot \frac{1}{180} = \frac{7\sqrt{10}}{75}</math>, so <math>p+q+r = \boxed{92}</math>. |
+ | |||
+ | NOTE: SIMPLY USE STEWARTS THEOREM | ||
==See Also== | ==See Also== |
Latest revision as of 06:48, 21 December 2020
Problem
Triangle is scalene. Points and are on segment with between and such that , , and . If and trisect , then can be written uniquely as , where and are relatively prime positive integers and is a positive integer not divisible by the square of any prime. Determine .
Solution
Let and . Since , by the Angle Bisector Theorem, we have and .
By using the Law of Cosines on and , we have
By using the Law of Cosines on and , we have
Multiplying the second equation by and adding the two equations results in
After substituting back, solve for to get
Thus, , so .
NOTE: SIMPLY USE STEWARTS THEOREM
See Also
2006 iTest (Problems) | ||
Preceded by: Problem 31 |
Followed by: Problem 33 | |
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