Difference between revisions of "2002 AMC 10B Problems/Problem 22"

Latest revision as of 16:28, 29 July 2011

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-== Problem ==
+#REDIRECT[[2002 AMC 12B Problems/Problem 20]]
-Let <math>\triangle{XOY}</math> be a right-triangle with <math>m\angle{XOY}=90^\circ</math>. Let <math>M</math> and <math>N</math> be the midpoints of the legs <math>OX</math> and <math>OY</math>, respectively. Given <math>XN=19</math> and <math>YM=22</math>, find <math>XY</math>.
-== Solution ==
-Let <math>OM=MX=x</math> and <math>ON=NY=y</math>. By the Pythagorean Theorem, <cmath>x^2+4y^2=484</cmath> and <cmath>4x^2+y^2=361</cmath> We wish to find <math>\sqrt{4x^2+4y^2}</math>. So, we add the two equations, multiply by <math>\frac{4}{5}</math>, and take the squareroot to get <math>XY=26</math>