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

(→‎Solution 4)
m (Fixed typos)
 
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By the Pythagorean Theorem again, we have  
 
By the Pythagorean Theorem again, we have  
  
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<cmath>(2x)^2 + (2y)^2 = XY^2 \Longrightarrow XY = \sqrt{4(x^2 + y^2)} = \sqrt{4(169)} = \sqrt{676} = \boxed{\mathrm{(B)}\ 26}</cmath>
+
<cmath>(2x)^2 + (2y)^2 = XY^2 \Longrightarrow XY = \sqrt{4(x^2 + y^2)} = \sqrt{4(169)} = \sqrt{676} = \boxed{\mathrm{(B)}\ 26}.</cmath>
  
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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>.
  
  
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== Solution 3 (Solution 1 but shorter) ==
 
== Solution 3 (Solution 1 but shorter) ==
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Refer to the diagram in solution 1. <math>4x^2+y^2=361</math> and <math>4y^2+x^2=484</math>, so add them: <math>5x^2+5y^2=845</math> and divide by 5: <math>x^2+y^2=169</math> so <math>\dfrac{XY}{2}=\sqrt{169}=13</math> and so <math>XY=26</math>, or answer <math>B</math>.
+
Refer to the diagram in solution 1. <math>4x^2+y^2=361</math> and <math>4y^2+x^2=484</math>, so add them: <math>5x^2+5y^2=845</math> and divide by 5: <math>x^2+y^2=169</math> so <math>\dfrac{XY}{2}=\sqrt{169}=13</math> and so <math>XY=26</math>, or answer <math>\boxed{\text{(B)}}</math>.
  
 
== Solution 4 ==
 
== Solution 4 ==
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Use the diagram in solution 1. Get <math>4x^2+y^2=361</math> and <math>4y^2+x^2=484</math>, and multiply the second equation by 4 to get <math>4x^2+16y^2=1936</math> and then subtract the first from the second. Get <math>15y^2=1575</math> and <math>y^2=105</math>. Repeat for the other variable to get <math>15x^2=960</math> and <math>x^2=64</math>. Now XY is equal to the square root of four times these quantities, so <math>(105+64) \cdot 4=676</math>, and <math>XY=26</math>
+
Use the diagram in solution 1. Get <math>4x^2+y^2=361</math> and <math>4y^2+x^2=484</math>, and multiply the second equation by 4 to get <math>4x^2+16y^2=1936</math> and then subtract the first from the second. Get <math>15y^2=1575</math> and <math>y^2=105</math>. Repeat for the other variable to get <math>15x^2=960</math> and <math>x^2=64</math>. Now XY is equal to the square root of four times these quantities, so <math>(105+64) \cdot 4=676</math>, and <math>XY=\boxed{\text{(B)}\ 26}</math>.
  
−
==Solution 5 (Quick and good for in - contest use)==
+
==Solution 5 (Quick and good for in-contest use)==
  
−
Let <math>XM = MO = y</math>, and <math>ON = NY  = x</math>. Now, we can write some equations. We have that <cmath>(2y)^2+x^2 = 19^2 \Longrightarrow 4y^2 + x^2 = 361</cmath>, and <cmath>(2x)^2+y^2 = 22^2 \Longrightarrow 4x^2+y^2 = 484</cmath>. Adding the <math>2</math> equations up we get <cmath>5x^2+5y^2 = 845</cmath>. The hypotenuse of the triangle is equal to <cmath>\sqrt{4x^2+4y^2}</cmath>, so that is what we want. In order to get <cmath>4x^2+4y^2</cmath>, we need to subtract <math>x^2+y^2</math> from <math>5x^2+5y^2</math>, and since <math>x^2+y^2 = \dfrac{1}{5} \cdot 5x^2+5y^2</math>. we have that <cmath>4x^2+4y^2 = \dfrac{4}{5} \cdot 845 \Longrightarrow 676</cmath>. Therefore, the hypotenuse is <math>\sqrt{676} = \boxed{\text{(B)}24}</math>
+
Let <math>XM = MO = y</math>, and <math>ON = NY  = x</math>. Now, we can write some equations. We have that <cmath>(2y)^2+x^2 = 19^2 \Longrightarrow 4y^2 + x^2 = 361,</cmath> and <cmath>(2x)^2+y^2 = 22^2 \Longrightarrow 4x^2+y^2 = 484.</cmath> Adding the <math>2</math> equations up we get <cmath>5x^2+5y^2 = 845.</cmath> The hypotenuse of the triangle is equal to <cmath>\sqrt{4x^2+4y^2},</cmath> so that is what we want. In order to get <cmath>4x^2+4y^2,</cmath> we need to subtract <math>x^2+y^2</math> from <math>5x^2+5y^2</math>, and since <math>x^2+y^2 = \dfrac{1}{5}\left(5x^2+5y^2\right)</math>, we have that <cmath>4x^2+4y^2 = \dfrac{4}{5} \cdot 845 = 676.</cmath> Therefore, the hypotenuse is <math>\sqrt{676} = \boxed{\text{(B)}\ 26}</math>.
  
 
-jb2015007
 
-jb2015007

Latest revision as of 12:24, 13 June 2025

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 1

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 $XY=2(MN)=\boxed{\mathrm{(B)}\ 26}$.


Solution 2

Let $XO=x$ and $YO=y.$ Then, $XY=\sqrt{x^2+y^2}.$

Since $XN=19$ and $YM=22,$ \[XN^2=19^2=x^2+(\dfrac{y}{2})^2)=\dfrac{x^2}{4}+y^2\] \[YM^2=22^2=(\dfrac{x}{2})^2+y^2=\dfrac{x^2}{4}+y^2.\]

Adding these up: \[19^2+22^2=\dfrac{4x^2+y^2}{4}+\dfrac{x^2+4y^2}{4}\] \[845=\dfrac{5x^2+5y^2}{4}\] \[3380=5x^2+5y^2\] \[676=x^2+y^2.\]

Then, we substitute: $XY=\sqrt{x^2+y^2}=\sqrt{676}=\boxed{26}.$


Solution 3 (Solution 1 but shorter)

Refer to the diagram in solution 1. $4x^2+y^2=361$ and $4y^2+x^2=484$, so add them: $5x^2+5y^2=845$ and divide by 5: $x^2+y^2=169$ so $\dfrac{XY}{2}=\sqrt{169}=13$ and so $XY=26$, or answer $\boxed{\text{(B)}}$.

Solution 4

Use the diagram in solution 1. Get $4x^2+y^2=361$ and $4y^2+x^2=484$, and multiply the second equation by 4 to get $4x^2+16y^2=1936$ and then subtract the first from the second. Get $15y^2=1575$ and $y^2=105$. Repeat for the other variable to get $15x^2=960$ and $x^2=64$. Now XY is equal to the square root of four times these quantities, so $(105+64) \cdot 4=676$, and $XY=\boxed{\text{(B)}\ 26}$.

Solution 5 (Quick and good for in-contest use)

Let $XM = MO = y$, and $ON = NY = x$. Now, we can write some equations. We have that \[(2y)^2+x^2 = 19^2 \Longrightarrow 4y^2 + x^2 = 361,\] and \[(2x)^2+y^2 = 22^2 \Longrightarrow 4x^2+y^2 = 484.\] Adding the $2$ equations up we get \[5x^2+5y^2 = 845.\] The hypotenuse of the triangle is equal to \[\sqrt{4x^2+4y^2},\] so that is what we want. In order to get \[4x^2+4y^2,\] we need to subtract $x^2+y^2$ from $5x^2+5y^2$, and since $x^2+y^2 = \dfrac{1}{5}\left(5x^2+5y^2\right)$, we have that \[4x^2+4y^2 = \dfrac{4}{5} \cdot 845 = 676.\] Therefore, the hypotenuse is $\sqrt{676} = \boxed{\text{(B)}\ 26}$.

-jb2015007

Video Solution by OmegaLearn

https://youtu.be/BIyhEjVp0iM?t=174

~ pi_is_3.14

Video Solution

https://www.youtube.com/watch?v=7wj6RupkO90 ~David

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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