Difference between revisions of "2008 AMC 12B Problems/Problem 17"
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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>. | ||
− | 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> <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>. |
This means that <math>\angle C=90^\circ</math> and <math>AC</math> is perpendicular to <math>BC</math>. So the slope of <math>BC</math> is the negative reciprocal of the slope of <math>AC</math>, yielding <math>m+n=\frac{1}{m-n}</math> <math>\Rightarrow</math> <math>m^2-n^2=1</math>. | This means that <math>\angle C=90^\circ</math> and <math>AC</math> is perpendicular to <math>BC</math>. So the slope of <math>BC</math> is the negative reciprocal of the slope of <math>AC</math>, yielding <math>m+n=\frac{1}{m-n}</math> <math>\Rightarrow</math> <math>m^2-n^2=1</math>. |
Revision as of 01:36, 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 , 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 .