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

Revision as of 03:35, 3 September 2021

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. $4x^2+y^2=19$ and $4y^2+x^2=22$ , 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 $B$ .

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 41: / Line 41: @@
 Then, we substitute: <math>XY=\sqrt{x^2+y^2}=\sqrt{676}=\boxed{26}.</math>
+== Solution 3 (Solution 1 but shorter) ==
+Refer to the diagram in solution 1. <math>4x^2+y^2=19</math> and <math>4y^2+x^2=22</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>.
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