Difference between revisions of "2008 Mock ARML 1 Problems/Problem 6"
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<cmath>\frac{AP}{NP} = \frac{AM}{MN} = \sqrt{\frac{5}{2}} \Longrightarrow NP = \sqrt{\frac{2}{5}}AP.</cmath> | <cmath>\frac{AP}{NP} = \frac{AM}{MN} = \sqrt{\frac{5}{2}} \Longrightarrow NP = \sqrt{\frac{2}{5}}AP.</cmath> | ||
Substituting, | Substituting, | ||
− | < | + | <cmath>\begin{align*} |
− | AP &= \frac{8\sqrt{5} \pm \sqrt{20}}{6} = \boxed{\frac{5\sqrt{5}}{3}}, \sqrt{5}.</ | + | 0 &= 3AP^2 - 8\sqrt{5}AP + 25\ |
+ | AP &= \frac{8\sqrt{5} \pm \sqrt{20}}{6} = \boxed{\frac{5\sqrt{5}}{3}}, \sqrt{5}. | ||
+ | \end{align*} | ||
+ | </cmath> | ||
== See also == | == See also == |
Latest revision as of 21:51, 13 April 2015
Problem
Square has side length . is the midpoint of , and is the midpoint of . is on such that is between and , and . Compute the length of .
Solution
By the Pythagorean Theorem, and . Let . By the Law of Cosines on , The Law of Cosines on yields
The Angle Bisector Theorem on yields Substituting,
See also
2008 Mock ARML 1 (Problems, Source) | ||
Preceded by Problem 5 |
Followed by Problem 7 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 |