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Difference between revisions of "2002 AMC 12B Problems/Problem 20"

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{{duplicate|[[2002 AMC 12B Problems|2002 AMC 12B #20]] and [[2002 AMC 10B Problems|2002 AMC 10B #22]]}}
 
{{duplicate|[[2002 AMC 12B Problems|2002 AMC 12B #20]] and [[2002 AMC 10B Problems|2002 AMC 10B #22]]}}
 
== Problem ==
 
== Problem ==
−
Let <math>\triangle XOY</math> be a [[right triangle|right-angled triangle]] with <math>m\angle XOY = 90^{\circ}</math>. Let <math>M</math> and <math>N</math> be the [[midpoint]]s of legs <math>OX</math> and <math>OY</math>, respectively. Given that <math>XN = 19</math> and <math>YM = 22</math>, find <math>XY</math>.
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Let <math>\triangle XOY</math> be a [[right triangle|right-angled triangle]] with <math>m\angle XOY = 90^{\circ}</math>. Let <math>M</math> and <math>N</math> be the [[midpoint]]s of legs <math>OX</math> and <math>OY</math>, respectively. Given that <math>XN = 19</math> and <math>YM = 22</math>, find <math>XY</math>.  
  
 
<math>\mathrm{(A)}\ 24
 
<math>\mathrm{(A)}\ 24
Line 8: Line 8:
 
\qquad\mathrm{(D)}\ 30
 
\qquad\mathrm{(D)}\ 30
 
\qquad\mathrm{(E)}\ 32</math>
 
\qquad\mathrm{(E)}\ 32</math>
 +
 
== Solution ==
 
== Solution ==
 
[[Image:2002_12B_AMC-20.png]]
 
[[Image:2002_12B_AMC-20.png]]

Revision as of 14:15, 21 June 2018

The following problem is from both the 2002 AMC 12B #20 and 2002 AMC 10B #22, so both problems redirect to this page.

Problem

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

2002 12B AMC-20.png

Let $OM = x$, $ON = y$. By the Pythagorean Theorem on $\triangle XON, MOY$ respectively, \begin{align*} (2x)^2 + y^2 &= 19^2\\ x^2 + (2y)^2 &= 22^2\end{align*}

Summing these gives $5x^2 + 5y^2 = 845 \Longrightarrow x^2 + y^2 = 169$.

By the Pythagorean Theorem again, we have

\[(2x)^2 + (2y)^2 = XY^2 \Longrightarrow XY = \sqrt{4(x^2 + y^2)} = \sqrt{4(169)} = \sqrt{676} = \boxed{\mathrm{(B)}\ 26}\]

Alternatively, we could note that since we found $x^2 + y^2 = 169$, segment $MN=13$. Right triangles $\triangle MON$ and $\triangle XOY$ are similar by Leg-Leg with a ratio of $\frac{1}{2}$, so $XY=2(MN)=\boxed{\mathrm{(B)}\ 26}$

There is the solution, folks! Overall, this problem's topic is associated with the Pythagorean theorem. If you do not understand this solution, you should take a look at everything about Pythagorean theorem. Please contact us if there are any questions, concerns, or doubts upon this problem, 

        Thank you.

See also

2002 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 21 		Followed by
Problem 23
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions
2002 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 19 		Followed by
Problem 21
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

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