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2002 AMC 12B Problems/Problem 20: Difference between revisions

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\qquad\mathrm{(E)}\ 32</math>
\qquad\mathrm{(E)}\ 32</math>


== Solution ==
== Solution ==  
[[Image:2002_12B_AMC-20.png]]
[[Image:2002_12B_AMC-20.png]]


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


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,  
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 ad triangles. Please contact us if there are any questions, concerns, or doubts about this problem, and we will take a closer look at our solution.
        Thank you.
      I hope you have a wonderful day!  Thank you.


== See also ==
== See also ==

Revision as of 14:24, 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

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 ad triangles. Please contact us if there are any questions, concerns, or doubts about this problem, and we will take a closer look at our solution. 

      I hope you have a wonderful day!  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

These problems are copyrighted © by the Mathematical Association of America.

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