Difference between revisions of "2008 AMC 12B Problems/Problem 17"
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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>. | ||
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+ | <math> \textbf{(A)}\ 16\qquad\textbf{(B)}\ 17\qquad\textbf{(C)}\ 18\qquad\textbf{(D)}\ 19\qquad\textbf{(E)}\ 20 </math> | ||
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Supposing <math>\angle A=90^\circ</math>, <math>AC</math> is perpendicular to <math>AB</math> and, it follows, to the <math>x</math>-axis, making <math>AB</math> a segment of the line x=m. But that would mean that the coordinates of <math>C</math> are <math>(m, m^2)</math>, contradicting the given that points <math>A</math> and <math>C</math> are distinct. So <math>\angle A</math> is not <math>90^\circ</math>. By a similar logic, neither is <math>\angle B</math>. | Supposing <math>\angle A=90^\circ</math>, <math>AC</math> is perpendicular to <math>AB</math> and, it follows, to the <math>x</math>-axis, making <math>AB</math> a segment of the line x=m. But that would mean that the coordinates of <math>C</math> are <math>(m, m^2)</math>, contradicting the given that points <math>A</math> and <math>C</math> are distinct. So <math>\angle A</math> is not <math>90^\circ</math>. By a similar logic, neither is <math>\angle B</math>. |
Revision as of 17:04, 29 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 ,
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
are
, 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
.