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

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== Problem ==
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== Problem 22 ==
  
 
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>.
 
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>.
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<math> \mathrm{(A) \ } 24\qquad \mathrm{(B) \ } 26\qquad \mathrm{(C) \ } 28\qquad \mathrm{(D) \ } 30\qquad \mathrm{(E) \ } 32 </math>
  
 
== Solution ==
 
== Solution ==
  
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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>
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Let <math>OM=MX=x</math> and <math>ON=NY=y</math>. By the Pythagorean Theorem, <math>x^2+4y^2=484</math> and <math>4x^2+y^2=361</math> 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 square root.
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<cmath>\begin{align*}
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5x^2+5y^2&=845\\
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4x^2+4y^2&=676\\
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\sqrt{4x^2+4y^2}&=\boxed{\mathrm{(B) \ } 26}
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\end{align*}</cmath>

Revision as of 14:46, 4 June 2011

Problem 22

Let $\triangle{XOY}$ be a right-triangle with $m\angle{XOY}=90^\circ$. Let $M$ and $N$ be the midpoints of the legs $OX$ and $OY$, respectively. Given $XN=19$ and $YM=22$, find $XY$.

$\mathrm{(A) \ } 24\qquad \mathrm{(B) \ } 26\qquad \mathrm{(C) \ } 28\qquad \mathrm{(D) \ } 30\qquad \mathrm{(E) \ } 32$

Solution

Let $OM=MX=x$ and $ON=NY=y$. By the Pythagorean Theorem, $x^2+4y^2=484$ and $4x^2+y^2=361$ We wish to find $\sqrt{4x^2+4y^2}$. So, we add the two equations, multiply by $\frac{4}{5}$, and take the square root. \begin{align*} 5x^2+5y^2&=845\\ 4x^2+4y^2&=676\\ \sqrt{4x^2+4y^2}&=\boxed{\mathrm{(B) \ } 26} \end{align*}

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