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>. | ||
+ | ==Alternate Solution== | ||
+ | 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: | ||
+ | <cmath>4x_A^2 + 7x_A - 1 + 4x_B + 7x_B - 1 = 0</cmath> | ||
+ | <cmath>4(x_A^2 + x_B^2) + 7(x_A + x_B) = 2</cmath> | ||
+ | <cmath>x_A^2 + x_B^2 = \frac{1}{2}.</cmath> | ||
+ | |||
+ | It also follows that <math>(x_A + x_B)^2 = 0</math>. Expanding this, we find: | ||
+ | <cmath>x_A^2 + 2x_A x_B + x_B^2 = 0</cmath> | ||
+ | <cmath>\frac{1}{2} + 2x_A x_B = 0</cmath> | ||
+ | <cmath>x_A x_B = -\frac{1}{4}.</cmath> | ||
+ | |||
+ | 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>: | ||
+ | <cmath>y_A - y_B = 4x_A^2 + 7x_A - 1 - 4x_B^2 - 7x_B + 1</cmath> | ||
+ | <cmath>= 4(x_A^2 - x_B^2) + 7(x_A - x_B)</cmath> | ||
+ | <cmath>= 4(x_A + x_B)(x_A - x_B) + 7(x_A - x_B)</cmath> | ||
+ | <cmath>= 7(x_A - x_B)</cmath> | ||
+ | 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}} |
Revision as of 18:21, 30 June 2013
Problem
Points and
are on the parabola
, and the origin is the midpoint of
. What is the length of
?
Solution
Let the coordinates of be
. As
lies on the parabola, we have
.
As the origin is the midpoint of
, the coordinates of
are
.
We need to choose
so that
will lie on the parabola as well. In other words, we need
.
Substituting for , we get:
.
This simplifies to , which solves to
. Both roots lead to the same pair of points:
and
. Their distance is
.
Alternate Solution
Let the coordinates of and
be
and
, respectively. Since the median of the points lies on the origin,
and expanding
, we find:
It also follows that . Expanding this, we find:
To find the distance between the points, must be found. Expanding
:
we find the distance to be
. Expanding this yields
.
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 |