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Difference between revisions of "2004 AMC 12B Problems/Problem 18"

(New page: == Problem == Points <math>A</math> and <math>B</math> are on the parabola <math>y=4x^2+7x-1</math>, and the origin is the midpoint of <math>AB</math>. What is the length of <math>AB</mat...)
 
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This simplifies to <math>8x_A^2 - 2 = 0</math>, which solves to <math>x_A = \pm 1/2</math>. Both roots lead to the same pair of points: <math>(1/2,7/2)</math> and <math>(-1/2,-7/2)</math>. Their distance is <math>\sqrt{ 1^2 + 7^2 } = \sqrt{50} = \boxed{5\sqrt2}</math>.
 
This simplifies to <math>8x_A^2 - 2 = 0</math>, which solves to <math>x_A = \pm 1/2</math>. Both roots lead to the same pair of points: <math>(1/2,7/2)</math> and <math>(-1/2,-7/2)</math>. Their distance is <math>\sqrt{ 1^2 + 7^2 } = \sqrt{50} = \boxed{5\sqrt2}</math>.
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==Alternate Solution==
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Let the coordinates of <math>A</math> and <math>B</math> be <math>(x_A, y_A)</math> and <math>(x_B, y_B)</math>, respectively. Since the median of the points lies on the origin, <math>x_A + x_B = y_A + y_B = 0</math> and expanding <math>y_A + y_B</math>, we find:
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<cmath>4x_A^2 + 7x_A - 1 + 4x_B^2 + 7x_B - 1 = 0</cmath>
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<cmath>4(x_A^2 + x_B^2) + 7(x_A + x_B) = 2</cmath>
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<cmath>x_A^2 + x_B^2 = \frac{1}{2}.</cmath>
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It also follows that <math>(x_A + x_B)^2 = 0</math>. Expanding this, we find:
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<cmath>x_A^2 + 2x_A x_B + x_B^2 = 0</cmath>
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<cmath>\frac{1}{2} + 2x_A x_B = 0</cmath>
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<cmath>x_A x_B = -\frac{1}{4}.</cmath>
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To find the distance between the points, <math>\sqrt{(x_A - x_B)^2 + (y_A - y_B)^2}</math> must be found. Expanding <math>y_A - y_B</math>:
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<cmath>y_A - y_B = 4x_A^2 + 7x_A - 1 - 4x_B^2 - 7x_B + 1</cmath>
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<cmath>= 4(x_A^2 - x_B^2) + 7(x_A - x_B)</cmath>
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<cmath>= 4(x_A + x_B)(x_A - x_B) + 7(x_A - x_B)</cmath>
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<cmath>= 7(x_A - x_B)</cmath>
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we find the distance to be <math>\sqrt{50(x_A - x_B)^2}</math>. Expanding this yields <math>5\sqrt{2(x_A^2 + x_B^2 - 2x_A x_B)} = \boxed{5\sqrt{2}}</math>.
  
 
== See Also ==
 
== See Also ==
  
 
{{AMC12 box|year=2004|ab=B|num-b=17|num-a=19}}
 
{{AMC12 box|year=2004|ab=B|num-b=17|num-a=19}}
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{{MAA Notice}}

Latest revision as of 15:29, 27 August 2016

Problem

Points $A$ and $B$ are on the parabola $y=4x^2+7x-1$, and the origin is the midpoint of $AB$. What is the length of $AB$?

$\mathrm{(A)}\ 2\sqrt5 \qquad \mathrm{(B)}\ 5+\frac{\sqrt2}{2} \qquad \mathrm{(C)}\ 5+\sqrt2 \qquad \mathrm{(D)}\ 7 \qquad \mathrm{(E)}\ 5\sqrt2$

Solution

Let the coordinates of $A$ be $(x_A,y_A)$. As $A$ lies on the parabola, we have $y_A=4x_A^2+7x_A-1$. As the origin is the midpoint of $AB$, the coordinates of $B$ are $(-x_A,-y_A)$. We need to choose $x_A$ so that $B$ will lie on the parabola as well. In other words, we need $-y_A = 4(-x_A)^2 + 7(-x_A) - 1$.

Substituting for $y_A$, we get: $-4x_A^2 - 7x_A + 1 = 4(-x_A)^2 + 7(-x_A) - 1$.

This simplifies to $8x_A^2 - 2 = 0$, which solves to $x_A = \pm 1/2$. Both roots lead to the same pair of points: $(1/2,7/2)$ and $(-1/2,-7/2)$. Their distance is $\sqrt{ 1^2 + 7^2 } = \sqrt{50} = \boxed{5\sqrt2}$.

Alternate Solution

Let the coordinates of $A$ and $B$ be $(x_A, y_A)$ and $(x_B, y_B)$, respectively. Since the median of the points lies on the origin, $x_A + x_B = y_A + y_B = 0$ and expanding $y_A + y_B$, we find: \[4x_A^2 + 7x_A - 1 + 4x_B^2 + 7x_B - 1 = 0\] \[4(x_A^2 + x_B^2) + 7(x_A + x_B) = 2\] \[x_A^2 + x_B^2 = \frac{1}{2}.\]

It also follows that $(x_A + x_B)^2 = 0$. Expanding this, we find: \[x_A^2 + 2x_A x_B + x_B^2 = 0\] \[\frac{1}{2} + 2x_A x_B = 0\] \[x_A x_B = -\frac{1}{4}.\]

To find the distance between the points, $\sqrt{(x_A - x_B)^2 + (y_A - y_B)^2}$ must be found. Expanding $y_A - y_B$: \[y_A - y_B = 4x_A^2 + 7x_A - 1 - 4x_B^2 - 7x_B + 1\] \[= 4(x_A^2 - x_B^2) + 7(x_A - x_B)\] \[= 4(x_A + x_B)(x_A - x_B) + 7(x_A - x_B)\] \[= 7(x_A - x_B)\] we find the distance to be $\sqrt{50(x_A - x_B)^2}$. Expanding this yields $5\sqrt{2(x_A^2 + x_B^2 - 2x_A x_B)} = \boxed{5\sqrt{2}}$.

See Also

2004 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 17 		Followed by
Problem 19
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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