Difference between revisions of "2008 AMC 12B Problems/Problem 17"
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− | Let the coordinates of <math>A</math> be <math>(m, m^2)</math> and the coordinates of <math>C</math> be <math>(n, n^2)</math>. Since the line <math>AB</math> is parallel to the <math>x</math>-axis, the coordinates of <math>B</math> must | + | Let the coordinates of <math>A</math> be <math>(m, m^2)</math> and the coordinates of <math>C</math> be <math>(n, n^2)</math>. Since the line <math>AB</math> is parallel to the <math>x</math>-axis, the coordinates of <math>B</math> must be <math>(-m, m^2)</math>. |
Then the slope of line <math>AC</math> is <math>\frac{m^2-n^2}{m-n}=\frac{(m+n)(m-n)}{m-n}=m+n</math>. | Then the slope of line <math>AC</math> is <math>\frac{m^2-n^2}{m-n}=\frac{(m+n)(m-n)}{m-n}=m+n</math>. | ||
The slope of line <math>BC</math> is <math>\frac{m^2-n^2}{-m-n}=-\frac{(m+n)(m-n)}{m+n}=-(m-n)</math>. | The slope of line <math>BC</math> is <math>\frac{m^2-n^2}{-m-n}=-\frac{(m+n)(m-n)}{m+n}=-(m-n)</math>. |
Revision as of 02:34, 23 December 2009
Let the coordinates of be
and the coordinates of
be
. Since the line
is parallel to the
-axis, the coordinates of
must be
.
Then the slope of line
is
.
The slope of line
is
.
Supposing , then
is perpendicular to
and, it follows, to the
-axis, making
a segment of the line x=m. But that would mean that the coordinates of
, contradicting the given that points
and
are distinct. So
is not
. By a similar logic, neither is
.
This means that and
is perpendicular to
. So the slope of
is the negative reciprocal of the slope of
, yielding
.
Because is the length of the altitude of triangle
from
, and
is the length of
, the area of
. Since
,
.
Substituting,
, whose digits sum to
.