2002 AMC 12B Problems/Problem 20

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

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

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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