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 22:51, 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
![[asy] pointpen=black;pathpen=black+linewidth(0.7); pair A=(0,2), D=(0,0), C=(2,0), B=(2,2), M=(C+D)/2, N=(B+C)/2, P=8/3*(N-M)+M; D(MP("A",A,(0,1))--MP("B",B,(0,1))--MP("C",C)--MP("D",D)--cycle); D(D(MP("M",M))--D(MP("N",N,E))--D(MP("P",P,E))); D(M--A--N--A--P); D(anglemark(M,A,P)); MP("2",(A+D)/2,W); [/asy]](http://latex.artofproblemsolving.com/a/e/8/ae872bd2d686fdfdf276cd736253bbd959031274.png)
By the Pythagorean Theorem, and
. Let
. By the Law of Cosines on
,
The Law of Cosines on
yields
![$NP^2 = AP^2 + \left(\sqrt{5}\right)^2 - 2 \cdot AP \cdot \left(\sqrt{5}\right) \cos \theta = AP^2 - \frac{8}{\sqrt{5}}AP + 5.$](http://latex.artofproblemsolving.com/8/b/f/8bfea20208ef683f90e102b9ac826d0b473ddeb9.png)
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 |