2009 AIME II Problems/Problem 3
Problem
In rectangle , . Let be the midpoint of . Given that line and line are perpendicular, find the greatest integer less than .
Solution
Solution 1
From the problem, and triangle is a right triangle. As is a rectangle, triangles , and are also right triangles. By , , and , so . This gives . and , so , or , so , or , so the answer is .
Solution 2
Let be the ratio of to . On the coordinate plane, plot , , , and . Then . Furthermore, the slope of is and the slope of is . They are perpendicular, so they multiply to , that is, which implies that or . Therefore so .
Solution 3
Similarly to Solution 2, let the positive x-axis be in the direction of ray and let the positive y-axis be in the direction of ray . Thus, the vector and the vector are perpendicular and thus have a dot product of 0. Thus, calculating the dot product:
Substituting AD/2 for x:
See Also
2009 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 2 |
Followed by Problem 4 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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