Difference between revisions of "2004 AMC 12A Problems/Problem 12"
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The [[equation]] of <math>\overline{AA'}</math> can be found using points <math>A, C</math> to be <math>y - 9 = \left(\frac{9-8}{0-2}\right)(x - 0) \Longrightarrow y = -\frac{1}{2}x + 9</math>. Similarily, <math>\overline{BB'}</math> has the equation <math>y - 12 = \left(\frac{12-8}{0-2}\right)(x-0) \Longrightarrow y = -2x + 12</math>. These two equations intersect the line <math>y=x</math> at <math>(6,6)</math> and <math>(4,4)</math>. Using the [[distance formula]] or <math>45-45-90</math> [[right triangle]]s, the answer is <math>2\sqrt{2}\ \mathrm{(B)}</math>. | The [[equation]] of <math>\overline{AA'}</math> can be found using points <math>A, C</math> to be <math>y - 9 = \left(\frac{9-8}{0-2}\right)(x - 0) \Longrightarrow y = -\frac{1}{2}x + 9</math>. Similarily, <math>\overline{BB'}</math> has the equation <math>y - 12 = \left(\frac{12-8}{0-2}\right)(x-0) \Longrightarrow y = -2x + 12</math>. These two equations intersect the line <math>y=x</math> at <math>(6,6)</math> and <math>(4,4)</math>. Using the [[distance formula]] or <math>45-45-90</math> [[right triangle]]s, the answer is <math>2\sqrt{2}\ \mathrm{(B)}</math>. | ||
Revision as of 19:21, 3 December 2007
Problem
Let 
and 
. Points 
and 
are on the line 
, and 
and 
intersect at 
. What is the length of 
?
Solution
The equation of can be found using points 
to be 
. Similarily, has the equation 
. These two equations intersect the line 
at 
and 
. Using the distance formula or 
right triangles, the answer is 
.
See also
| 2004 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 11 |
Followed by Problem 13 |
| 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 | |
